# Full text of "Advanced algebra"

## See other formats

```I.-^

Digitized by the Internet Archive

in 2008 with funding from

IVIicrosoft Corporation

BY

HEEBEET E. HAWKES, Ph.D.

Assistant Professor of Mathematics in Yale University

X

GINN & COMPANY

BOSTON . NEW YORK • CHICAGO • LONDON

QA\54

e.

H. E. HAWKES

ALL BIGHTS KESERVED

613.1

GINN & COMPANY • PRO-
PRIETORS . BOSTON . U.S.A.

PEEFACE

This book is designed for use in secondary schools and in
short college courses. It aims to present in concise but clear
form the portions of algebra that are required for entrance to
the most exacting colleges and teclinical schools.

The chapters on algebra to quadratics are intended for a
review of the subject, and contain many points of view that
should be presented to a student after he has taken a first
course on those topics. Throughout the book the attention
is concentrated on subjects that are most vital, pedagogically
and practically, while topics that demand a knowledge of the
calculus for their complete comprehension (as multiple roots,
and Sturm's theorem) or are more closely related to other por-
tions of mathematics (as theory of numbers, and series) have
been omitted.

The chapter on graphical r epresentation, has been intro-
duced early, in the belief that the illumination which it affords
greatly enlivens the entire presentation of algebra. The dis-
cussion of the relation between , pairs of linear ftguntioT^ ^ g and
pairs of straight lines is particularly su g gestive. _

In each chapter the discussion is directed toward a definite

result. The chapter on theory of equations aims to give a

simple and clear treatment of the method of obtaining the

real roots of an equation and the theorems that lead to that

iii

383513

vi CONTENTS

CHAPTER n
FACTORING

SBOnOK ^ PAGE

28. Statement of the Problem 16

29. Monomial Factors 16

30. Factoring by grouping Terms 17

31. Factors of a Quadratic Trinomial 18

32. Factoring the Difference of Squares 20

33. Reduction to the Difference of Squares 20

34. Replacing a Parenthesis by a Letter 21

35. Factoring Binomials of the Form a" ± 6« 22

36. Highest Common Factor 22

S'7. H.C.F. of Two Polynomials 23

i^.:)i}adid"s Method of finding the H.C.F 23

39. Method of finding the H.C.F. of Two Polynomials ... 24

40. Least y^o>nmon Multiple 26

41. Second Rule for finding the Least Common Multiple ... 26

CHAPTER III
FRACTIONS

42. General Principles . 27

43. Principle I 27

44. Principle II 27

45. Principle III 27

46. Reduction. . 27

47. Least Common Denominators of Several Fractions ... 28

48. Addition of Fractions 29

49. Subtraction of Fractions 29

60. Multiplication of Fractions 29

51. Division of Fractions 29

CHAPTER IV
EQUATIONS

52. Introduction . 32

63. Identities and Equations of Condition 32

64. Linear Equations in One Variable 33

65. Solution of Problems . . 37

66. Linear Equations in Two Variables 40

57. Solution of a Pair of Equations 40

CONTENTS

vu

SECTION

68. Independent Equations

69. Solution of a Pair of Simultaneous Linear Equations

60. Incompatible Equations

61. R^um6

62. Solution of Problems involving Two Unknowns

63. Solution of Linear Equations in Several Variables

PAGE
. 41

• 42
. 42

43
• 45

47

CHAPTER V
RATIO AND PROPORTION

64. Ratio 49

65. Proportion . 49

66. Theorems concerning Proportion 49

67. Theorem v 50

68. Mean Proportion 60

CHAPTER VI

IRRATIONAL NUMBERS AND RADICALS

69. Existence of Irrational Numbers .

70. The Practical Necessity for Irrational Numbers

71. Extraction of Square Root of Polynomials .

72. Extraction of Square Root of Numbers .

73. Approximation of Irrational Numbers .

74. Sequences

75. Operations on Irrational Numbere

76. Notation

77. Other Irrational Numbers ....

78. Reduction of a Radical to its Simplest Form

79. Addition and Subtraction of Radicals .

80. Multiplication and Division of Radicals .

81. Rationalization

82. Solution of Equations involving Radicals

62
53
53
54
55
56
56
57
57
68
69
60
61
63

CHAPTER VII
THEORY OF INDICES

83. Negative Exponents 66

84. Fractional Exponents 66

85. Further Assumptions 67

86. Theorem 67

87. Operations with Radical Polynomials 69

viii CONTENTS

i CHAPTER VIII

SECTION PAGE

88. Definition 70

89. Solution of Quadratic Equations 70

90. Pure Quadratics 72

91. Solution of Quadratic Equations by Factoring .... 75

92. Solution of an Equation by Factoring 75

93. Quadratic Form 77

94. Problems solvable by Quadratic Equations 79

95. Theorems regarding Quadratic Equations 82

96. Theorem 83

97. Theorem 84

98. Nature of the Roots of a Quadratic Equation 84

CHAPTER IX
GRAPHICAL REPRESENTATION

99. Representation of Points on a Line 87

100. Cartesian Coordinates 88

101. The Graph of an Equation 90

102. Restriction to Coordinates 91

103. Plotting Equations 91

104. Plotting Equations after Solution 93

105. Graph of the Linear Equation . "94

106. Method of plotting a Line from its Equation .... 96

107. Solution of Linear Equations, and the Intersection of their Graphs 97

108. Graphs of Dependent Equations 99

109. Incompatible Equations 09

110. Graph of the Quadratic Equation 100

111. Form of the Graph of a Quadratic Equation .... 101

112. The Special Quadratic ox^ + to = 103

113. The Special Quadratic aa;^ + c = 104

114. Degeneration of the Quadratic Equation 104

115. Sum and Difference of Roots 106

116. Variation in Sign of a Quadratic 107

CONTENTS ix

CHAPTER X
SIMULTANEOUS QUADRATIC EQUATIONS IN TWO VARIABLES

SECTION PAGE

117. Solution of Simultaneous Quadratics Ill

118. Solution by Substitution Ill

119. Number of Solutions 113

120. Solution when neither Equation is Linear 114

121. Equivalence of Pairs of Equations 120

122. Incompatible Equations 121

123. Graphical Representation of Simultaneous Quadratic Equations 122

124. Graphical Meaning of Homogeneous Equations . . . 123

CHAPTER XI
MATHEMATICAL INDUCTION

126. General Statement 125

CHAPTER XII
BINOMIAL THEOREM

126. Statement of the Binomial Theorem 128

127. Proof of the Binomial Theorem 129

128. General Term 129

CHAPTER XIII

ARITHMETICAL PROGRESSION

129. Definitions : .... 133

130. The nth Term 133

131. The Sum of the Series 134

132. Arithmetical Means 134

CHAPTER XIV

GEOMETRICAL PROGRESSION

133. Definitions 137

134. The nth Term 137

135. The Sum of the Series 138

136. Geometrical Means 138

137. Infinite Series 140

CONTENTS

CHAPTER XV

PERMUTATIONS AND COMBINATIONS
SECTION PAGE

138. Introduction 143

139. Permutations 144

140. Combinations 146

141. Circular Permutations 149

142. Theorem 160

CHAPTER XVI

COMPLEX NUMBERS

143. The Imaginary Unit 152

144. Addition and Subtraction of Imaginary Numbers . . . 163"

145. Multiplication and Division of Imaginaries 154

146. Complex Numbers 156

147. Graphical Representation of Complex Numbers .... 155

148. Equality of Complex Numbers . 155

149. Addition and Subtraction 156

150. Graphical Representation of Addition 156

151. Multiplication of Complex Numbers 157

152. Conjugate Complex Numbers . . . . . . . 158

153. Division of Complex Numbers 158

154. Polar Representation 160

155. Multiplication in Polar Form 160

156. Powers of Numbers in Polar Form 161

157. Division in Polar Form 162

158. Roots of Complex Numbers 162

CHAPTER XVII

THEORY OF EQUATIONS

159. Equation of the nth Degree 166

160. Remainder Theorem 166

161. Synthetic Division . . 167

162. Proof of the Rule for Synthetic Division 169

163. Plotting of Equations . . 170

164. Extent of the Table of Values 171

^

CONTENTS xi

SIECTION PAGE

fl65. Roots of an Equation 172

. Number of Roots 172

167. Graphical Interpretation 174

^ 168. Imaginary Roots . . . 174

. Graphical Interpretation of Imaginary Roots .... 175

170. Relation between Roots and Coefficients 177

171. The General Term in the Binomial Expansion .... 178

172. Solution by Trial . 178

173. Properties of Binomial Surds 179

174. Formation of Equations . . 180

176. To multiply the Roots by a Constant 183

176. Descartes' Rule of Signs 186

177. Negative Roots 189

178. Integral Roots 190

179. Rational Roots 190

180. Diminishing the Roots of an Equation . . ... . 191

181. Graphical Interpretation of Decreasing Roots .... 193

182. Location Principle 194

183. Approximate Calculation of Roots by Horner's Method . . 195

184. Roots nearly Equal 200

CHAPTER XVIII

DETERMINANTS

185. Solution of Two Linear Equations 203

186. Solution of Three Linear Equations 204

187. Inversion 208

188. Development of the Determinant 208

189. Number of Terms 210

190. Development by Minors 210

191. Multiplication by a Constant 213

192. Interchange of Rows and Columns 213

193. Interchange of Rows or Columns 214

194. Identical Rows or Columns . . . . . . . 215

195. Proof for Development by Minors 215

196. Sum of Determinants 216

197. Vanishing of a Determinant 217

198. Evaluation by Factoring 218

199. Practical Directions for evaluating Determinants . ... 219

200. Solution of Linear Equations 221 ^

201. Solution of Homogeneous Linear Equations .... 223 4^^ — -•

xii CONTENTS

CHAPTER XIX

PARTIAL FRACTIONS
SECTION .^ PAGE

202. Introduction 226

203. Development when (f){x)= has no Multiple Roots . . . 225

204. Development when (p{x)=0 has Imaginary Roots . . . 229

205. Development when <p (x) = {x — ar)» 232

206. General Case 233

CHAPTER XX
LOGARITHMS

207. Generalized Powers 235

208. Logarithms . ' 236

209. Operations on Logarithms 237

210. Common System of Logarithms 239

211. Use of Tables 241

212. Interpolation 242

213. Antilogarithms 243

214. Cologarithms 246

215. Change of Base 247

216. Exponential Equations 261

217. Compound Interest 263

CHAPTER XXI
CONTINUED FRACTIONS

218. Definitions . ' 256

219. Terminating Continued Fractions 256

220. Convergents 258

221. Recurring Continued Fractions 260

222. Expression of a Surd as a Recurring Continued Fraction . 263

223. Properties of Convergents 266

224. Limit of Error 267

CHAPTER XXII
INEQUALITIES

225. General Theorems . .269

226. Conditional Linear Inequalities 271

227. Conditional Quadratic Inequalities 271

CONTENTS xiii

CHAPTER XXm
VARIATION

SECTION ~ PAOB

228. General Principles . .273

CHAPTER XXIV
PROBABILITY

229. Illustration 276

230. General Statement 276

CHAPTER XXV
SCALES OF NOTATION

231. General Statement 279

232. Fundamental Operations ... .... 280

233. Change of Scale ..... .... 281

234. Fractions 282

236. Duodecimals . , 284

ik

CHAPTER I
FUNDAMENTAL OPERATIONS

1. It is assumed that the elementary operations and the mean-
ing of the -usual symbols of algebra are familiar and do not
demand detailed treatment. In the following brief exposition
of the formal laws of algebra most of the proofs are omitted.

2. Addition. The process of adding two positive integers a and
h consists in finding a number x such that

a -\-h = x.

For any two given positive integers a single sum x exists
which is itself a positive integer.

3. Subtraction. The process of subtracting the positive num-
ber h from the positive number a consists in finding a number x

such that

h -\- x = a. (1)

This number x is called the difference between a and h and is

denoted as follows :

a — 6 = a;,

a being called the minuend and h the subtrahend.

If a > ft and both are positive integers, then a single posi-
tive integer x exists which satisfies the condition expressed by
equation (1)

1

. AL'G^BE a: to * ^JJ ADR ATI€S ^

li a <h, then x is not a positive integer. In order that the pro-
cess of subtraction may be possible in this case also, we introduce
negative numbers which we symbolize by (—a ), (— b), etc. When
in the difference a~b,ais less than b, we define a — b = (— (b — a)).
The processes of addition and subtraction for the negative numbers
are defined as follows :

a -^ (— b) = a — b.

a — (— b) = a -^ b.

4. Zero. If in equation (1), a = b, there is no positive or nega-
tive number which satisfies the equation. In order that in this
case also the equation may have a number satisfying it, we intro-
duce the number zero which is symbolized by and defined by

the equation

a -\- = a,

or a — a = 0.

The processes of addition and subtraction for this new number
zero are defined as follows, where a stands for either a positive or
a negative number

0-a = -a.
± = 0.

5. Multiplication. The process of multiplying a by i consists
in finding a number x which satisfies the equation

a-b = X.

* The symbol for a positive integer might be written (+ a), (+ 6), etc., consistently
with the notation for negative numbers. Since, however, no ambiguity results, we omit
the + sign. Since the laws of combining the + and — signs given in this and the folloM'ing
paragraphs remove the necessity for the parentheses in the notation for the negative
number, we shall omit them where no ambiguity results.

FUNDAMENTAL OPERATIONS 3

When a and h are positive integers cc is a positive integer which,
may be found by adding a to itseK h times. When the numbers
to be multiplied are negative we have the following laws,

(— a) • (— 5) = a • 5,

Oa = «;.0 = 0, (1)

where «; is a positive or negative number or zero.

These symbolical statements include the statement of the
following

Principle. A 'product of numbers is zero when and only when
one or more of the factors are zero.

This most important fact, which we shall use continually, assures
us that when we have a product of several numbers as

a-b-G- d =^ Bj

first, if e equals zero, it is certain that one or more of the num-
bers a, bj c, or d are zero ; second, if one or more of the numbers
a, b, c, or d are zero, then e is also zero.

6. Division. The process of dividing ahj p consists in finding
a number x which satisfies the equation

X-p = a, . (1)

where a and jS are positive or negative integers, or a is 0.
When a occurs in the sequence of numbers

• ••-.?A -2A -Ap, ft-2A 3A •••,

a; is a definite integer or 0, that is, it is a number such as we
have previously considered. If a is not found in this series, but
is between two numbers of the series, then in order that in this
case the process may also be possible we introduce the fraction

which we symbolize by a -*- /8 or - and which is defined by the
equation

4 ALGEBRA TO QUADRATICS

The operations for addition, subtraction, multiplication, and divi-
sion of fractions are defined as follows :

fi 8 Py W

Further properties of fractions are the following :

a 1

— = — > where S is any number, (5)

j8 SP

The last two equations are expressed verbally as follows :

Both numerator and denominator of a fraction may be multi-
plied by any number without changing the value of the fraction.

Changing the sign of either, numerator or denominator of a
fraction is equivalent to changing the sign of the fraction.

The laws of signs in multiplication given on p. 3 may now be
assumed to hold when the letters represent fractions as well as
integers. j- ,.-, .

Thus for example ~ I t ) ' "~

ac
M

The positive or negative number a may be written in the

fractional form

a

1*

7. Division by zero. If in equation (1), ^Q>^ p=i 0, there is no
single number x which satisfies the equation, since by (1), § 5,
whatever value x might have, its product with zero would
be zero.

FUNDAMENTAL OPERATIONS 5

Thus division by zero is entirely excluded from algebraic pro-
cesses. Before a division can safely be performed one must be
assured that the divisor cannot vanish. In the equation

4 = 2.0,

if we should allow division of both sides of our equation by zero,
we should be led to the absurd result 4 = 2.

8. Fundamental operations. The operations of addition, sub-
traction, multiplication, and division we call the four fundamental
operations. Any numbers that can be derived from the number 1
by means of the four fundamental operations we call rational num-
bers. Such numbers comprise all positive and negative integers
and such fractions as have integers for numerator and denominator.
Positive or negative integers are called integral numbers.

9. Practical demaild for negative and fractional numbers.
In the preceding discussion negative numbers and fractions have
been introduced on account of the mathematical necessity for
them. They were needed to make the four fundamental opera-
tions always possible. That this mathematical necessity corre-
sponds to a practiqal necessity appears as soon as we attempt
to apply our four operations to practical affairs. Thus if on a
certain day the temperature is + 20° and the next day the mer-
cury falls 25°, in order to express the second temperature we
must subtract 2^ from 20. If we had not introduced negative
numbers, this would be impossible and our mathematics would
be inapplicable to this and countless other everyday occurrences.

10. Laws of operation. All the numbers which we use in
algebra are subject to the following laws.

' .Commutative law of addition. This law asserts that the value
of the sum of two numbers does not depend on the order of
summation.

Symbolically expressed,

a -\- h = h -\- aj

where a and h represent any numbers such as we have presented
or shall hereafter introduce.

6 ALGEBRA TO QUADRATICS

Associative law of addition. This law asserts that the sum of
three numbers does not depend on the way in which the numbers
are grouped in performing the process of addition.

Symbolically expressed,

a 4- (^ + c) = (a + &) + c = a 4- * + c.

Commutative law of multiplication. This law asserts that the
value of the product of two numbers does not depend on the order
of multiplication.

Symbolically expressed,

a-b = b-a.

Associative law of multiplication. This law asserts that the
value of the product of three numbers does not depend on the way
in which the numbers are grouped in the process of multiplication.

Symbolically expressed,

(^a-b) -c = a-(b'C)= a-b-c.

Distributive law. This law asserts that the product of a single
number and the sum of two numbers is identical with the sum
of the products of the first number and the other two numbers
taken singly.

Symbolically expressed,

a- (b -\- c)= a-b -{- a-c.

All the above laws are readily seen to hold when more than three numbers are
involved.

11. Integral and rational expressions. A polynomial is inte-
gral when it may be expressed by a succession of literal terms, no
one of which contains any letter in the denominator.

Thus 4 a;5 _ a;3 _ 2 x2 _ J x + 1 is integral.

The quotient of two integral expressions is called rational.

^^ a;2_2a; + 3. ^. ,
Thus z — 18 rational.

a — 7

12. Operations on polynomials. We assume that the same
formal laws for the four fundamental operations enunciated in
§ § 2-6 and the laws given in § 10 hold whether the letters in the
symbolic statements represent numbers or polynomials.

FUNDAMENTAL OPERATIONS 7

In fact the literal expressions which we use are in essence
nothing else than numerical expressions, since the letters are merely
symbols for numbers. When the letters are replaced by numbers,
the literal expressions reduce to numerical expressions for which
the previous laws have been explicitly given.

13. Addition of polynomials. For performing this operation
we have the following

EuLE. Write the terms with the same literal part in a column.
Find the algebraic sum of the terms in each column, and write
the results in succession with their proper signs.

When the polynomials reduce to monomials the same rule is to be observed.

EXERCISES

Add the following :

1. 3a262_2a6 + 6a26-a; 4. ab - 2 a^h'^ - 11 a'^b + 9 a ;

2a^b -ab -2a; Sam- 4:db -6a.

Solution: ZaW-2ab+ 6a^b- a

-2a262 + 4a6- lla26 + 9a

- a6 + 2 a26 - 2 a

Sa^-iab -6a

9am -Sab- Sa^b

2. 21a-246-8c2; 16c + 17 6 + 6c2 - 20a ; 186 -18c.

3. x* -6a;2- 8x-l; 2x3 + 1; 6x'^ + 1 x + 2 ; x^ - x^ + x - 1.

4. 9(a + 6)-6(6 + c) + 7(a + c); 4(6 + c) - 7(a + 6) - 8(a + c) ,•

(a + c) - (a + 6) + (6 + c).

5. a2-4a6 + 62 + a + 6-2; 2a2 + 4a6- 362- 2a -26 + 4;

3a2- 5a6-462 + 3a + 36 - 2; 6a2 + 10a6 + 562 + a + 6.

14. Subtraction of polynomials. For performing this opera-
tion we have the following

EuLE. Write the subtrahend under the minuend so that terms
with the same literal part are in the same column.

To each term of the minuend add the corresponding term of
the subtrahend, the sign of the latter having been changed.

It is generally preferable to imagine the signs of the subtrahend changed rather
than actually to write it with the changed signs.

8 ALGEBRA TO QUADRATICS

EXERCISES

1. From a262 - 3 a^ft + 8 a6 + 6 6 subtract 9 a'^h'^ - 6 a6 + 4 a^ft + a.

Solution : a%'^ - 3 a26 + 8 a& + 6 6

-9a262:f 4a26- 6a6 +a

- 8a262 ^ 7 a26 + 14a6 + 66 - a

2. From 6 ahx — 4 win + 5 x subtract 3 mn + 6 ax — 4 a6x.

3. From m -{■ an + bq subtract the sum of

cm + dn + {b — a)q and {a — b)q — {a -\- d)n — cm.

4. From the sum of | a + y^ 6 + | c and — 6 — c — a subtract ^6 — |c + ^a.

5. From the sum of 2 x^ — 3 x + 4 and x* — f x — i subtract x^ — | x^

-3|x + 3i.

15. Parentheses. When it is desirable to consider as a single
symbol an expression involving several numbers or symbols for
numbers, the expression is inclosed in a parenthesis. This paren-
thesis may then be used in operations as if it were a single number
or symbol, as in fact it is, excepting that the operations inside
the parenthesis may not yet have been carried out.

EuLE. When a single parenthesis is preceded hy a -\- sign
the parenthesis may he removed, the various terms retaining
the same sign.

When a single parenthesis is preceded hy a — sign the parent-
thesis may be removed, providing we change the signs of all the
terms inside the parenthesis.

When several parentheses occur in an expression we have the
following

EuLE. Remove the innermost parenthesis, changing the signs
of the terms inside if the sign preceding it is minus.

Simplify, if possible, the expression inside the new inn^rwjost
parenthesis.

Repeat the process until all the parentheses are removed.

It is in general unwise to shorten the process by carrying out some of the steps
in one's head. The liability to error in such attempts more than offsets the gain
in time.

FUNDAMENTAL OPERATIONS

EXERCISES

Remove parentheses from the following :

1. 6 _ pa -[26 + (4a -2a -6) -66]}.

Solution: 6 - {9a - [26 + (4a - 2a - 6) - 66]}

= 6- {9a -[26 + (4a -2a + 6) -66]}
= 6-{9a-[2 6+ (2a + 6)-6 6]}
= 6 _ [9 a - (2 6 + 2 a + 6 - 6 6)]
= 6- [9a -(2a -36)]
= 6 -(9a -2a + 36)
= 6 - (7 a + 3 6)
= 6-7a-36
= - 7 a - 2 6.

2- -{-[-[-(-(-I))]]}-

3. a2 + 4 - {6 - [- (a2 - 6) + 1]}.

4. i{i-|[f-ia-i-f-T\) + |]-^}-

5. x2 - {2/2 _ [4x + 3(y - 9x(y - x)) + 9y (x - y)]}.

6. Find the value of a - {5 6 - [a - (3 c - 3 6) + 2 c - 3 (a - 26 -c)]} i^
when a = — 3, 6 = 4, c = — 5.

16. Multiplication. It is customary to write a - a = a^ ;
aaa = a^] a - a • • a = a"*. We have tlien by tlie associative

n terms

law of multiplication, § 10,

a^ - a^ = (a • a) (a - a ' a) = a a • a ' a • a = a^,
or, in general.

where m and n are positive integers.
Furthermore,

(I)

(a')" = a"-a'---a' = a'-"'. (II)

m terms

Finally, a" • J" = (a • bf. (Ill)

The distinction between (a**)"* and a""* should be noted carefully. Thus (28)*
= 82 = 64, while 28* =2^ = 512.

Equation (I) asserts that the exponent in the product of two
powers of any expression is the sum of the exponents of the
factors. Hence we may multiply monomials as follows :

10 ALGEBRA TO QUADRATICS

I

Rule. Write the product of the numerical coefficients, followed ,
by all the letters that occur in the multiplier and multiplicand,
each having as its exponent the sum of the exponents of that
letter in the multiplier and multiplicand.

Example. 4 a2^i0c# • (- 16 a'^bd'^) = - 64 a^b^^cd^\

17. Multiplication of monomials by polynomials. By the

distributive law, § 10, we can immediately formulate the

EuLE. Multiply each term of the polynomial by the monomial
and write the resulting terms in succession.

Example. 9 a'^h'^ - 2 a& + 4 a62 - a + &*

3a26

27 a463 _ 6 aW + 12 a^h^ - 3 a^fe + 3 aW
*

18. Multiplication of polynomials. If in the expression for the

distributive law, a(c + d)= ac •{- ad,

we replace a by a + 5, we have

(a -\- h) (c -\- d) = ac -\- be -\- ad -{- bd,
which affords the

EuLE. Multiply the mult^icand by each term of the multi-
plier in turn, and write the partial products in succession.

To test the accuracy of the result assume some convenient numer-
ical value for each letter, and find the corresponding numerical
value of multiplier, multiplicand, and product. The latter should
be the product of the two former.

EXERCISES

1. Multiply and check the following:

(a) 2 a2 + a6 + 4 62 + 5 and a-h-\- ah.

Check :

Solution :

Let a = 6 = 1

2a2+ a6+462 + 6

= 8

a - h + a6

= 1

2a8+ a26 + 4a&2 + a6

-2a26- a62 -463_

-62

+ a62

+ 2a86 + a262 + 4a68

2a8 - a26 + 4a62 + a6 - 46^ - 62 -f- 2a86 + a262 + 4a68 = 8

FUNDAMENTAL OPERATIONS H

(b) 6 abx^ and 4 a'^b^x.

(c) ^^and -OxV^.

o

(d) 3 a62x - ^ 6x4 and 6 a^cx.

(e) x2« + ?/26 ^ x^ and x« — y^

(f ) a^ + ab + l^ and a^ + ac + c2.

(g) x^y^'^j a;»-3yin4-4^ and ic^y2m-2^
(h) xP-3 4- xP-2 + 1 and a;3 _ a;2 _ 1,

(i) 8 a26c, - a62, _ 7 62, - — a^c*, and -.
w ' 4 ' ' 14 ' 6

(j) ax* — 2 a2x3 — X + 4 a and — x + 2 a.
(k) x« + * + x2« + x2& 4- a;3o-& and x«-^ — 1.

(1) 15x* - 11x3 + 6x2 + 2x - 1 and - 3x2 - 1.
(m) 4 x* — 8 xy^ + | x2y2 _ 3 x^y — x — y and — 42 xy.

2. Expand (x + y)*.

3. Expand and simplify

(a;2 + 2/2 + 22)2 _ (X + y + 2r) (X + y - z) (x + 2 - y) (y + 2 - X).

19. Types of multiplication. The following types of multi-
plication should be so familiar as merely to require inspection of
the factors in order to write the product.

EuLE. The product of the sum and difference of two terms is
equal to the square of the terms with like signs minus the square
of the terms which have unlike signs.

Examples. (a - 6) (a + 6) = a?^ — 62.

(4x2 - 3y2)(4x2 + 3y2) = 16x4 - 9y4.

20. The square of a binomial. This process is performed as
follows :

KuLE. The square of a binomial, or expression in two terms,
is equal to the sum of the squares of the two terms plus twice
their product.

Examples. (x + y)2 = x2 + y2 + 2 xy.

(2a-36)2=:4a2 + 962- 12a6.

12 ALGEBRA TO QUADRATICS

21. The square of a polynomial. This process is performe!
as follows :

Rule. The square of any polynomial is equal to the sum
of the squares of the terms plus twice the product of each term
hy each term that follows it in the polynomial.

Example, (a + 6 + c)2 = a^ + 62 + c^ + 2 a6 + 2 ac + 2 6c.

22. The cube of a binomial. This process is performed as
follows :

Rule. The cube of any binomial is given hy the folloimng

expression : ^^ _^ ^y ^ ^s _^ ^ ^2j -\.3ab^ + h\

EXERCISES

Perform the following processes by inspection.
1. (a-6 + c)2. 2. (a4-66)2.

3. (2x'-i-l)2 4. (a2-62)3.

5. (2x'-i-l)3. 6. (l-8a;22/)2.

7. (X2-2X + 1)2. 8. (a;2 - y2 + ^2)2.

9. (2a-26-c)2. 10. (x8-2x-l)2.

11. (ai'-3-6p + 3)2. 12. (-6x2y + 4xy2)8.

13. {xp - 2/9) {xp + y^). 14. (- 6x22/ + 4 xy^)^.

15. (3x+ 2 2/)(3x-2 2/). 16. (- 3ax2 + 2ax -6)2.

17. {-3x^y-^lz^){Sx'^ + ^z^). 18. (_ 4 - 6a26) (- 4 + 6a26).

19. (2 a -2^. 20. (2 a-?)'.

23. Division. By the definition of division in § 6, we have

a = —} a'^ = — ) a^ = —z'y
. a a a''

or, in general, ^

a"-™ = — ->
a""

where n and m are positive integers and n> m.

If n = m, we preserve the same principle and write

a«-» = a« = — = 1,

(1)

FUNDAMENTAL OPERATIONS 13

24. Division of monomials. Keeping in mind the rule of signs
for division given in § 6, we have the following

EuLE. Divide the numerical coefficient of the dividend hy
that of the divisor for the numerical coefficient of the quotient^
keeping in mind the rule of signs for division.

Write the literal part of the dividend over that of the
divisor in the form of a fraction^ and perform on each pair
of letters occurring in both numerator and denominator the
process of division as defined hy equation (I) in the preceding
paragraph.

Example. Divide 12 a'^lA^cH by - 6 a^hc^d^.
12 a^6"c2d _ 2 bio

25. Division of a polynomial by a monomial. This process is
performed as follows :

Rule. Divide each term of the polynomial hy the monomial
and write the partial quotients in succession.
Example. Divide 8 a^lP - 12 a^h^ by 2 aW.

8a266 _ 12 gcftg _ 463 _ 6a»

2a363 2a363 ~ a & *

26. Division of a polynomial by a polynomial. This process
is performed as follows :

Rule. Arrange hoth dividend and divisor in descending
powers of some common letter {called the letter of arrangement)-

Divide the first term of the dividend hy the first term of the
divisor for the first term of the quotient.

Multiply the divisor hy this first term of the quotient and
subtract the product from the dividend.

Divide the first term of this remainder hy the first term of the
divisor for the second term of the quotient, and proceed as
before until the remainder vanishes or is of lower degree in the
letter of arrangement than the divisor.

14 ALGEBRA TO QUADRATICS

When the last remainder vanishes the dividend is exactly divisible by
divisor. This fact may be expressed as follows:

dividend _ . . .

,. . = quotient.

divisor

When the last remainder does not vanish we may express the result of division

dividend .. , , remainder

-^rr-. = quotient -\ -— :

divisor divisor

The coefficients in the quotient will be rational numbers if those in both divi-
dend and divisor are rational.

EXERCISES

Divide and check the following :

1. 8a8 + 6a26 + 9a62 + 963by4a + &.

Solution : 4a4-&| 8 a^ + 6 a26 + 9 a62 + 9 h^ \2 a^ + ab + 2lfl

8 a3 + 2 a'^b

4 a26 + 9 a62
4a26+ am

8a&2 + 9&3
8a&2 + 268
76»
7 63 . ■

Result : 2a^-\-ah-{-2,b^ +

1 4a + 6 •

Check : Let a = h =1. Dividend = 32, divisor = 5, quotient = 6§.
32 - 5 = 6f .

2. xi2 - yi2 by x3 - 2/3. 3. 2 x2 - 6x + 2 by X - 2.
4. xi2 — 2/12 i3y aj4 _ 2/*. 5. x^ — 2/6 by x^ + xy -\- y^.
6. a8 - a2 + 2 bya + 1. 7. - 63x*y32;2 ^y _ gx^y^z.

8. .x2 - X - 30 by X + 5. 9. 4 a26 - 6 a62 _ 2 a by - 2a. •

10. 16 a264cii by - 2 a^¥c\ 11. i x2 - 3^ x - | by 1^ x + T»ff.

12. ax2 + (a2 - 6) X - a6 by X + a.

13. ax« + 6x»-i + cx«-2 - dx«-3 by x'.

14. 16 a2x22/2 _ 8 ax32/2 - 4 x*?/ by - f xy.

15. ISai'b^ -\- 6aP + 2?^+3_ 9 ap + ^ft^ by 3 apft*.

16. a2 - 2 a6 - 4 c2 + 8 6c - 3 62 by a - 2 c + 6.

17. x* — (d + 6 + c) x2 -f {ab + ac + 6c) x - a6c by x - a.

18. 2j/2 _ 6x2 + i/xy + V-x - -^y + 1 by 2x + |y - f.

19. x8 - 2 x22/ - x2 + 2/2x f 2 xy - y - 2/2 + 2 by X - y + 1.

20. 3x8 + 6x22/ + 9x2 + 2x2/2 + 5^3 + 22/ + 6y2 + 3 by x + 22/ + 3.

FUNDAMENTAL OPERATIONS 15

27. Types of division. The following types of division, which
may be verified by the rule just given for any particular integral
value of rij should be so familiar that they may be performed by
inspection.

(a^n _ ^,2n) ^ (^^» -t ^n^ = «»» ^ ^«. (I)

(a" + b^) -^ (a + 5) = a"-i - a^-% + a^-%'' + b^-\ (II)

where n is odd.

(a" -. h^) ^(a-b)^ a"-i 4- a"-'^ + a«-3^>2 + . . . + j«-i^ (III)
where n is odd or even.

EXERCISES

Give by inspection the results of the following divisions.
1. a« - 1 by a -r l: 2. a^ + 1 by a + 1.

3. x' + 128 by ic + 2. 4. x^ + y^hy x-\- y.

6. x^ — y^ by x^ -{■ y^. 6. x^ — y* hj x — y.

7. x8 - ^8 by X* - y*. 8. a2m _ 1 by a - 1.

9. a2«+^ - 1 by a - 1. 10. 27a9 + 868 by 3^8 + 26.

11. 8x8-27 by 2a:- 3. 12. 4a2 - 25668 by 2a + 1664.

13. 16 a* - 256 by 4 a^ + 16. 14. 27 ai^ - 64 612 by 3 a* - 4 6*.

CHAPTER II
FACTORING

28. Statement of the problem. The operation of division con-
sists in finding the quotient when the dividend and divisor are
given. The product of the quotient and the divisor is the divi-
dend, and the quotient and the divisor are the factors of the
dividend. Thus the process of division consists in finding a
second factor of a given expression when one factor is given.

The process of factoring consists in finding all the factors of
a polynomial when no one of them is given. This operation is in
essence the reverse of the operation of multiplication. We shall
be concerned only with those factors that have rational coefficients.

29. Monomial factors. By the distributive law, § 10,

ah -{• ac = a(}) -\- c).
This affords immediately the

Rule. Write the largest monomial factor which occurs in every
term outside a jpareiithesis which includes the algebraic sum of
the remaining factors of the various terms.

EXERCISES

Factor the following :

1. 6 Q?Wc + 9 alPc^ - 15 a*6c7.

Solution : 6 a'^hH -t- 9 ah^c^ - 15 a^hc^ = 3 a6c (2 at^ -J- 3 ftScS - 5 a^<fi).

2. 14 anx — 21 6nic — 7 n.

3. 121 a'^hH - 22 a^hc'^ + H oJb'^cK

4. 5xV - 10x8?/8 - SxV - 15x2y2.

5. 21 ahn + 6 obH"^ - 18 a'^hv?- + 15 a'^h'^n.

6. 10 al^cmx - 5 ab^cy + 6 ab'^cz - 15 abc^m^.
7 a^xV - 49 ax^y* + 14 axy^z^ - 21 a^'^y^.

8. 45 a*62c8d - 9a6*c«d8 -I- 27 a^bc*d^ - 117 a^l^cd*.

16

FACTORING 17

30. Factoring by grouping terms. If in the expression for
the distributive law, ac + bc = {a-\- b) c,

we replace chj c -\- d,

we have a(G -\- d) -\- h (c -\- d) = ac -\- ad -\- be -\- bd.

We may then factor the right-hand member as follows :
ac_ -^ ad -\- be -\- bd = a (c -\- d) -\- b {c -\- d) = {a -{- b) (c -\- d).

This affords the

EuLE. Factor out any monomial expression that is common to
each term of the polynomial.

Arrange the terms of the polynomial to be factored in groups
of two or more terms each, such that in each group a monomial
factor may be taken outside a parenthesis which in each case
contains the same expression.

Write the algebraic sum of the monomial factors that occur
outside the various parentheses for one factor, and the expres-
sion inside the parentheses for the other factor.

EXERCISES

Factor the following :

1. 4a3&-6a262_4a4 + 6a63.

Solution : 4 a^b - 6 a'^h'^ - 4 a* + 6 aft*

= 2 a (2 a26 - 3 a62 _ 2 a3 + 3 63)

= 2 a (2 a26 - 2 a^ - 3 a62 + 3 &»)
= 2 a [2 a2 (6 ^a) + 3 62 (6 _ a)]
= 2 a (6 - a) (2 a2 + 3 62).

2. x2-(3a + 46)x + 12a6.

Solution : x2 - (3 a + 4 6) x + 12 a6

= x2 - 3 ax - 4 6x + 12 a6
= X (x — 3 a) — 4 6 (x — 3 a)
= (x-46)(x-3a).

3. 2 ax - 3 6y - 2 6x + 3 ay. 4. 56 a2 - 40 a6 + 63 ac - 45 6c.

5. a''x2 — 6'?y3 _ f/jx^ + apy\ 6. 91 x2 - 112 mx + 65 nx - 80 mru

7. ax - bx -[■ ex + ay - by -\- cy. 8. 2 ax - 6x - c6 + 2 a6 + 2 ac - 62.

18 ALGEBRA TO QUADRATICS

9. 2x6 - 3x8 + 2x« - 3. 10. x^ + x^ + x + 1.

11. acx"^ - hex + adx-bd. 12. 2 6x - x^ - 4 & + 2 x.

13. 3x3 _ 12x82/2 _ 4y2 + 1. 14. 4x2 - (8 + b)x + 2 6.

15. x* - (4m + 9n)x2 + 36 mn. 16. x* - (2m + 3n)x2 + 6mn.

17. a2x-ac+a6y-a62x-63y +c62. 18. 18 a» - 2 ac*&6 _ g a26 + c^b^.
19. 2ax — 6ay + a-26x + 56y-6. 20. 2ax-ay-26x + 4cx— 2cy+62/.

21. 2 a2x« + 4 a2x* + 2 a2x2 + 4 a^.

22. 8x2 - 2 ox - 12 xz + 3az + 4xy - ay.

31. Factors of a quadratic trinomial. In this case we cannot
factor by grouping terms immediately, as that method is inappli-
cable to a polynomial of less than four terms. We observe, how-
ever, that in the product of two binomial expressions.

(mx + n) (px + qy= tnp^J -h {mq + w^)ic -|- ngy

the coefficient of x is the sum of two expressions mq and np^

whose product is equal to the product of the coefficient of x^ and

the last term, that is,

mq • np = m,p • nq.

Thus, to factor the right-hand member of this equation, we may
remove the parenthesis from the term in x and use the principle
rf grouping terms. Thus

mpx^ -f (mq + np) q? + nq

= TYipx^ -j- m.qx + npx -f- nq
= mx (px -\- q)+ n (px -f q)
= (mx + n) (px -\- q).
This affords the

Rule. Write the trinomial in order of descending powers of
X (or the letter in which the expression is quadratic).

Multiply the coefficient of a^ by the term not involving x, arid
find two factors of this product whose algebraic sum is the
coefficient of x.

Replace the coefficient of x by this sum and factor by group-
ing terms.

FACTORING 19

Factoring a perfect square is evidently a special case under this method.
Thus factor x2 + 6x + 9.

1-9=9. 3 + 3 = 6.

x2 + (3 + 3)x + 9
= x2 + 3x + 3x + 9
= a;(x + 3) + 3(x + 3)
= (X + §) (X + 3)

= (X + 3)2.

One will usually recognize when a trinomial is a perfect square, in which
case the factors may be written down by inspection.

+ 2 = 8.

EXERCISES

Factor the following :

(

1. 3x2 + 8x + 4.

/

Solution :

3-4 =

12. / 6 H

3x2 + 8x + 4
= 3x2 + (6 + 2)x + 4
= 3x2 + 6x + 2x + 4
= 3x(x + 2) + 2(x + 2)
= (3x + 2)(x + 2).

2. 8x2-

146X + 352.

Solution :

8. 362:.. 2462.

8x2-146x + 362

26 -.126 = -146.

= 8x2-(2 6 + 12 6)x + 362
= 8x2-126x-26x + 362
= 4x(2x-36)-6(2x-36)
= (4x-6)(2x-36).
3. 28x2 -3x- 40.

Solution : 28 • (- 40) = - 1120.

The factors of 1120 must be factors of 28 and 40. We seek two factors
of 1120, one of which exceeds the other by 3. We note that since 40 exceeds
28 by more than 3, one factor must be greater and the other less than 28 and
40 respectively.

Since 4 • 7 = 28 and 5 ■ 8 = 40,

we try 5 • 7= 35 and 4 • 8 = 82,

which are the required factors of 1120.

28x2 -3x- 40
= 28x2-(35-32)x-40
= 28 x« - 35 X + 32 X - 40
= 7x(4x-6) + 8(4x-6)
= (7x + 8)(4x-5).

20 ALGEBRA TO QUADRATICS

4. x2-6x + 9. 5. 2x2 + re -6.

6. 2x2 - X - 6. 7. 2x2 + X - 91.

8. x2 + X - 182. 9. 9x2 - 2x - 7.

10. 2x2 + 5a; + 3. 11. x2 - llx + 18.

12. 3x2-10x-8. 13. 6x2 + 17 x + 7.

14. 16x2 + 4x- 3. 15. 7?/2-4?/-ll.

16. a2-6a6 + 962. 17. 5x2-12x + 4.

18. 18x2-73x + 4. 19. 27x2 + 3x- 2.

20. 24x2 - 31x - 16. 21. 21x2 - 31x + 4.

22. 12 x2 + 60 X - 72. 23. x* - 3 az^-A 2 a^.

24. 9x2-18ax-7a2. 25. 10x2-63x-13.

26. x4y« - 12 x2?/3 + 36. 27. 4x2p-16xi'- 81. >v

28. 2 x8 - 17 6x2 + sif'^x. 29. 4 a2 + 12 a6 + 9 62.

30. 6 a2x2 - 2 a6x - 7 62. ' 3 1. 10 x* - 16 a2x3 - 100 x^a^.

32. 4 aa"* + 16 a^b» + 16 62«. 33. 4 a^x^y* - 20 a6xy2z + 25 62z2.

32. Factoring the difference of squares. Under the method of
the preceding paragraph we may factor the difference of squares.

Thus to factor x^ — b^ we observe that the product of the
coefficient of x^ and the constant term is

1 . (_ b^) = _ b^.

Since the coefficient of x. is zero, we have

-b-\-b = 0.

Hence (x -\- b)(x — b)— x^ — b\

EuLE. Extract the square root of each term.
The sum of these square roots is one factor, and their differ-
ence is the other.

Example. Factor 9a2xV - 16 68c2.

9 a2x«2/* - 16 68c2 = (3 axV + 4 64c) (3 ax-3y2 _ 4 54c) .

33. Reduction to the difference of squares. The preceding
method may be used when the expression to be factored becomes
a perfect square by the addition of the square of some expression.

FACTORING 21

EXERCISES

Factor the following :

1.44 ^4 6*.

Solution r a* + 4 6* = a^ + 4 0,262 + 4 54 _ 4 ^^252

= (a2 + 2b2)2_4a262
= (a2 + 2 62 _ 2 ab) {a^'-\- 2 62 + 2 db).

2. 1 - a*. - 3. a* + 4.

4. x^ — X. 5. x«y4 + 4 x2.

6. 4a;4+.2/*. 7. 4a2-2562.

8. x* + ic2 ^. 1. 9 ig ^254 _ 3.4.

10. x4 + 9x2 + 81. 11. 4a2p - 962c2«.

12. a2p + 3 - 16a36*. 13. ar*« + x2« + 1.

14. 36x2^4^8 - 49 w2ui6. 15. x4 - 13x2 + 36.

16. nty^ X IG mx* - lij m'x^y'^. 1 7. 9 x* + 8 x^^ + 4 y^.

34. Replacing a parenthesis by a letter. Any of the preceding methods
may be applied when a polynomial appeare in place of a letter in the expres-
sion to be factored. It is frequently desirable for simplicity to replace such
a polynomial by a letter, and in the final result to restore the polynomial.

EXERCISES

Factor the following :

1. 2 ax2 - 2 6x2 - 6 ax + 6 6x - 8 a + 8 6.

Solution : 2 ax2 - 2 6x2 - 6 ax + 6 6x - 8 a + 8 6

= 2(a — 6)x2 - 6(a - 6)x - 8(a - 6)
= (a -6) (2x2 -6.x -8)
= 2{a-6)(x2-3x-4)
= 2 (a - 6) (X - 4) (X + 1).

In this example the factor (a — 6) might have been replaced by a letter.

2. a2 + 62 - c2 - 9 - 2 a6 + 6 c.

Solution : a2 + 62 - c2 - 9 - 2 a6 + 6c

= a2 - 2 a6 + 62 - (c? - 6c + .9)
= (a - 6)2 -^ (c - 3)2 .

= (a - 6 + c - 3) (a - 6 - c + g.

3. (3x-2/)(2a+p)-(3x-2/)(a-9).

4. (4a-66)(3m-2p) + (a + 56)(3m-2p).

5. (7a-32/)(5c-2d)-(6a-22/)(5c-2d).

6. (X - 2/) (3a + 46) - (4a - 56) (X - 2/) - (X - y) (2a - 86).

22 ALGEBRA TO QUADRATICS

7. 6(x + 2/)2-ll(x4-2/)-7.

8. 4a2 - 12a6 + 962 _ x2 - 2x - 1.

9. x2a2 + 2 x2a + x2 - a2 - 2 a - 1.
10. ax2 + 6 ox + 9a - &x2 - 66x - 96.

fll. 4 (a - 6)2 - 5(a2 - 62) - 21 (a + 6)2.

12. 6(x + ?/)2 - 12 (x2 -y2)^4{x- yf.

13. a262x2 - a262 - 2 a6x2 + 2 a6 + x2 - 1.

14. (x-2y)(2a-36)-(96-10)(x-2?/).

/ 35. Factoring binomials of the form d^ ± 6*^. By § 27,

b/ ^^ V ci" + ^" = (« + ^'Xa"-^ - a"-2^ + a«-3^»2 _|_ ^,«-i>)^

where w is odd.

One can factor by inspection any binomial of the given form
by reference to these equations.

EXERCISES

Factor the following :

1. x^ — 2/6.

Solution : x^ - ye - ^xB _ ^3) (a;3 + ^3)

= (X2 + Xy + 7/2) (X - y) (X2 - Xy + 2/2) (x _|. y).

2. x6 + 125. 3. x^ - 1.
4. xi2 - 2/12. 5. x^ — 2/9.

6. Xl8 - 2/18. 7. a;16 _ yl6.

8. a2x3 + a5. 9. x* - a82/4.

. 10. 216 a + a*. 11. ox* - 16 a.

12. 3a7 - 96 65a2. 1^-"" 13. 2,1 x^^ + 64 2/8. ^"^^

14. 27 x52/7 + x22/4. «--^ 15. 16a468-81ci6d8,/^

36. Highest common factor. An expression that is not further
divisible into factors with rational coefficients is called prime.

If two polynomials have the same expression as a factor, this
expression is said to be their common factor.

The product of the common prime factors of two polynomials
is called their highest common factor, or H.C.F.

The same common prime factor may occur more than once. Thus (x — 1) 2 (x + 1)
and (x - 1)2 (a; - 2)2 have (x -X^ as their H.C.F. v-^—

c^f

FACTORING 23

37. H.C.F. of two polynomials. The process of finding the
H.C.F. is performed as follows:

EuLE. Factor the polynomials. The product of the common
prime factors is their H.C.F.

EXERCISES [l^'^-Xi^\)i^^^UH

Find the H.C.F. of the following:

1. 4 ab^x^ - 8 ab^x^ + 4 ab^ and 6 abx"^ + 12 abx + 6ab.
Solution : — 4 ab'^x^ - 8 ab^^ + 4 ab^

= 4a62(x4- 2x2+1)
= 4a62(x-l)2(x + l)2.

6 a6x2 + 12 abx + 6ab
= 6a6(x2 + 2x + 1)
= 6a6(x + 1)2.

The H.C.F. is then 2 a6(x + 1)2.

2. x^ — y^ and x2 — y"^.

3. x3 + x2 - 12 X and x2 + 5x + 4.

4. 9mx2 — Qmx + m and 9nx2 — n.

5. 6x- 4x2 + 2 ax -3a and 9 - 4x2.

6. 12 a2 - 3(3a6 + 27 62 and 8a2 - 1862.

7. 3 a2x - 6 a6x + 3 62x and 4 02^/ - 4 62y.

8. 2 X - 46 - x2 - 2 6x and 4x - 5x2 - 6.

9. 6x^-7 ax2 - 20 a2x and 3 x2 + ax - 4 a2.

38. Euclid's method of finding the H.C.F. When one is unable to factor
the polynomials whose H.C.F. is sought, the problem may nevertheless be
solved by use of a method which in essence dates from Euclid (300 e.g.).

The validity of this process depends on the following
Principle. If a polynomial has a certain factor, any multiple of it has the
same factor.

Let x» + ^x»-i + 5x»-2 + . . . +£:

and . x« + ax'»-i +6x'»-2 H 1- ^

be represented by F and 6? respectively. The letters A, B, ■ ■, K and a, 6,

• • •, Z represent integers, and m, the degree of G, is no greater than n, the

degree of F. We seek a method of finding the H.C.F. of F and G if any

exists. Call Q the quotient obtained by dividing F by G, and call B the

remainder. Then (§ 26)

^' F=QG-^B, (1)

24 ALGEBRA TO QUADRATICS

where the degree of E in x is not so great as that of G. Now whatever the
H.C.F. of i^'and G may be, it must also be the H.C.F. of G and B. For since

F-QG = B,

the H.C.F. of F and G must be a factor of the left-hand member, and hence
a factor of R, which is equal to that member. Also every factor common
to G and R must be contained in F, for any factor of G and E is a factor of
the right-hand member of (1), and hence of F.

Thus our problem is reduced to finding the H. C. F. of G and R. Let Qi and
Bi be respectively the quotient and remainder obtained in dividing G by R.

Then (? = QiB 4- Ru

where the degree of Ei in x is not as great as that of B. By reasonirg simi-
lar to that just employed we see that the H.C.F. of G and R is also the
H.C.F. of B and Bi. Continue this process of division.

Let B = Q2R1 + R2,

-Bi = QsRi + Eg.
until, say in Rk = Qk+2Rk + \ + Bk+i,

either B^ is exactly divisible by B^ +1 (i.e. E^- + 2 = 0), or i?;t + 2 does not con-
tain X. This alternative must arise since the degrees in x of the successive
remainders E, Ei, E2, • • • are continually diminishing, and hence either the
remainder must finally vanish or cease to contain x. Suppose Bk + 2=0. Then
the H.C.F. of Bk and Bk + 1 is Ea.- + 1 itself, which must, by the reasoning given
above, be also the H.C.F. of F and G. If Rk + 2 does not contain x, then the
H.C.F. of F and G, which must also be a factor of Ea + 2, can contain no x,
and must therefore be a constant.

Thus F and G have no common factor involving x.

This process is valid if the coefficients of F and G are rational expressions in
any letters other than x.

39. Method of finding the H.C.F. of two polynomials. The above dis-
cussion we may express in the following

Rule. Divide the polynomial of higher degree {if the degrees of the polyno-
mials are unequal) by the other, and if there is a remainder, divide the divisor
by it ; if there is a remainder in this process, divide the previous remainder
by it, and so on until either there is no remainder or it does not contain the letter
of arrangement. If tJiere is no remainder in the last division, the last divisor is
the H. C. F. If the last remainder does not contain the letter of arrangement,
then the polynomials have no common factor involving that letter.

In the application of this rule any divisor or remainder may be multiplied or
divided by any expression not involving the letter of arrangement without affect-
ing the H.C.F.

FACTORING

?6

EXERCISES
Find the H.C.F. of the following:

2x

1. 2x* + 2x3-x2
Solution : /Q ^

Multiply by -

1 andx4 + x3 + 4x + 4.

2J2x* + 2x3-x2- 2x-l |xH x^+ 4x + 4

2x^ + 2x3 + 8x + 8

■1, -x2-10x-9 *■ ^' ^

x^-\-10x+9\ x^+ x3+ 4x + 4 |x2-9x + 81

Ri

x4 + 10x3+ 9x2

,- [^-«)

- 9x3- 9x2+ 4x ^t4H/l

- 9x3-^90x2- 81x I

Divide by - 725,

(9-!. c , ec,

x + 9| x2 + lOx + 9 |x + l \

81x2+ 85X+4
81x2 + 810x+729
- 726ig 7 a 6

X2 +

>y

/'te.F.

9x + 9

9x + 9

Thus the H.C.F. is x + 1.
This process may be performed in the following more compact form.

2

2x4 + 2x3-x2- 2x-l

a^+ x^+ 4x + 4

x2-9x + 81

2x* + 2x3 + 8x + 8
-x2 - lOx-9

X4 + I0x3+ 9X2

-1

- 9x^- 9x2+ 4x

x + 9

x2 + 10x + 9

- 9x3- 90x2- 81 X

X2+ X

81x2+ 85x + 4

9x+9
9x + 9

81x2+810x +729

-725x -725

-725

X +1

Result: x + 1.

2. x2 + 6 X - 7 and x^ - 39x + 70.

3. x^ — X* — X + 1 and 5 x* — 4 x^ — 1.

4. x8 + 2x2 + 9 and - 6x3 - iia;2 + i5x + 9.

5. x3 - 2x2 - 15x + 36 and 3x2 - 4x - 15.

6. x* - 3x3 + x2 + 3x - 2 and 4 x3 - 9x2 + 2 X + 3.

7. 4x8 - 18x2 + 19x - 3 and 2x* - 12x3 + 19x2 _ 6x + 9.

8. X* + 4x3 _ 22x2 - 4x + 21 and x* + 10x8 + 20x2 - lOx - 21.

9. 6 a*x3 - 9 a3x2y - 10 a^ xy^ + 16 ay^ and 10 a^xV - 1^ «*«^y* + 8 a^*y^
12 a2x3y6.

26 ALGEBRA TO QUADRATICS

40. Least common multiple. The least common multiple of two

or more polynomials is the polynomial of least degree that con-
tains them as factors. We may find the least common multiple
of several polynomials by the following

EuLE. Multiply together all the factors of the various poly-
nomials, giving to each factor the greatest exponent with which it
appears in any of the polynomials,

41. Second rule for finding the least common multiple. When
only two polynomials are considered the previous rule is evidently
equivalent to the following

EuLE. Multiply the polynomials together and divide the
product hy their highest common factor.

EXERCISES

Find the least common multiple of the following :

1. x2 - y2,a;2 + ay — ax - xy, and a;2 - 2 xy + y\

Solution : x^ -y'^ = {x - y)ix -^ y).

x'^ + ay -ax-xy = {x- y) {x — a).
x^-2xy + y^ = {x- yf.

Thus the L.C.M. = (x - yY (x + y) (x - a).

2. 4a26c, 6a62, and 12 c2

3. 9x?/2, 6x22/3, and ZxyH"^.

4. (X + 1) (x2 - 1) and x3 - 1.

5. x* + 4x22/2 and x2 + 2 2/2 - 2 y.

6. 4x2 - 9?/ and 4x2 - 12x2/ + 9?/2.

7. x2 - 4 X + 3, x2 - 1, and x2 - ax — x + a.

8. x-1, 2x2-5x-3, and2x8-7x2 + 2x + 3.

9. x* - 9x2 + 26x - 24 and x^ - 10x2 + 31 x - 30.

10. 2x2 - 3x - 9, x2 - 6x + 9, and 3x2 _ 9x - 6x + 3&.

CHAPTEE III
FRACTIONS

42. General principles. The symbolic statements of the rules
for the addition, subtraction, multiplication, and division of alge-
braic fractions are the same as the statements of the correspond-
ing operations on numerical fractions given in (2), (3), and (4),
§ 6. This is immediately evident if we keep in mind the fact
that algebraic expressions are symbols for numbers and that if the
letters are replaced by numbers, the algebraic fraction becomes a
nunierical fraction.

43. Principle I. Both numerator and denominator of a frac-
tion may he multiplied (or divided) hy the same expression with-
out changing the value of the fraction.

This follows from (5), § 6.

44. Principle II. If the signs of both numerator and denomi-
nator of a fraction he changed, the sign of the fraction remains
unchanged.

This follows from Principle I, when we multiply both numerator and
denominator by — 1 .

45. Principle III. If the sign of either numerator or denomi-
nator (hut not hoth) he changed, the sign of the fraction is changed.

This follows from (6), § 6.

46. Reduction. A fraction is said to be reduced to its lowest
terms when its numerator and denominator have no common
factor. We effect this reduction by the following

Eule. Divide hoth numerator and denominator hy their
highest common factor.

27

28 ALGEBRA TO QUADRATICS

EXERCISES
* Reduce the following to their lowest terms.
12 ax2 - 12 ab^

1.

4ax^-Sabx-\-4:al»

12ax2-12a62 12 a {x - b) {x ■{■ b)

4 ax2 - 8 a6x + 4 ab^

Solution :

H.C.F. =4a{x-b).

a;S + 4x *^ a — 2ax — lOx + 5

g 6x2 -8ax +2a2 2 a26 + 2 a62 - 2 abc

x2-a2 * ■ 3 6c2 - 3 62c - 3 abc '

g a2 + 62 _ c2 + 2 a& 21x8-9x2 + 7x - 3

xi8 - ai8* • a2 - 62 4. c2 + 2 ac' ' 3x8 + 15x2 + x + 5 *

^j^ 2x2+ 3x-9 j^2 X* - x8 - X + 1 j^„ x3 + 3ax2 + 3a2x + a8
x2-9 ' ' 2x4 -x3-2x + l' * a2 + 2ax + x2

47. Least common denominator of several fractions. We have
the following

Rule. Find the least common multiple of the various denomi-
nators.

Multiply both numerator and denominator of each fraction hy
the expression which will make the new denominator the least
common multiple of the denominators.

EXERCISES

Reduce the following to their least common denominator.

- 2 8 ^ 2x-3

1. -> , and — -.

X 2x-l 4x2-1

Solution : The L.C.M. of the denominators is x (4x2 — 1) . Thus the frac-
tions are

2 (4 x2 - 1) 3x(2x+l) ^^ x(2x-3)

x(4x2-l)' x(4x2-l) ' ^^ x(4x2-l)'
" , and 3. , ■ , and

6 + a a2-62 2x-8 4x2 + 4x-16 4x2-26

FRACTIONS 29

A 1 1 ^ 1

4. , , and

aj3 _ y3 aj4 _ y4 a;2 - y2

5. , — - — , and

c 2x-\ X ^ 2x-3

6. — , , and

x2 - 2 X + 1 x2 - 1 (x + 1)2

-a b . c

7. : , , and

a + b-c a-\-b -^c d^ + 2ab + b^ - c"^

48. Addition of fractions. This operation we perform as
follows :

EuLE. Reduce the fractions to he added to their least common
denominator.

Add the numerators for the numerator of the sum, and take
the least common denominator for its denominator.

49. Subtraction of fractions. This operation we perform as
follows :

Rule. Reduce the fractions to their least common denomi-
nator.

Subtract the numerator of the subtrahend from that of the
minuend for the numerator of the result, and take the least
common denominator for its denominator.

50. Multiplication of fractions. This operation we perform
as follows :

Rule. Multiply the numerators together for the numerator
of the product, and the denominators for its denominator.

51. Division of fractions. This operation we perform as
follows :

Rule. Invert the terms of the divisor and multiply by the
dividend. a

Remark. Since a fraction is a means of indicating division, t~^^ and -

are two expressions for the same thing.

d

30 ALGEBRA TO QUADRATICS

EXERCISES ^

Perform the indicated operations and bring the results into their simplest
forms.

a + b a — b

I I* — w a + b

' a + b a — b

a + &

Solution ;

a 4- 6 a — b
a ^b a + 6
a + b a — b
a—b a+b

a^-l^

{a + 6)2 + (a ■

-6)2

a2-62

(a + 6)2 _ (a .

-6)2

a2-62

(a + 6)2 + (a -

-6)2

a2-62

(

2 a2 + 2 62 a2 + 62

4a6

2a6

iW - 1

2.V *

3 + f + i

(a + 6)2 - (a - 6)2

' 1 + 1 ' 2.V * • 1-|-t\

5 2-V- g 3 + f + l 7 2 + f + f

* i-(-W ■f + l + i.V '1-l + f"

8. ^_^L±^. 9. ^-i^-^. 10. -i- +

6 26 ab ac be a — b a + b

11.1+1+1. 12. '?^l^+!-^ 13. 20=

a6c c c a-la2-l

-, 3a;-l 2«-7 -- 2ic-l 2x-5

14. . 15.

l-3x 7 a;-2a;-4

16.—?^ L_. 17. ^ 8

4x-4 6a; + 6 3x-9 6«-15

18 « + ^ ■ g'^ + 2 a6 - 62 2a-36 3a-26

' a-6 ■ a2-62 " " 12 a "*" 16a

20 ?. ■ Q 21 «t? + ^c acZ-6c

15(x-l) 10(x + l) 2cd{c-d) 2cd{c + d)

22 a? + y a;-y 4zy ^^ a;2 + a;(a + 6) + a6 gg-gg
' x-y x-\-y x2-y2' ' x2 - x(a + 6) + a6' x^ - 62'

24. i5 + il + i?£-l^. 25. ^ «

6 a 14 a 36 a 16 a a2 - 9 a + 14 ' a* - 6 a - 14

26. Z^ + lU/^-l + lV 27 g(«-'g) a(a + x)

\y8 a;/ \y2 y x) a2 + 2 ttx + x2 a2-2ax + xa*

28. lzi^.lz^./i + _±_V 29. -A_ + _^ + _A._l.
1 + 6 a + a« V 1-a/ (x - 1)« (x - 1)2 ^ x - 1 x

FRACTIONS 31

^ 1

30. a + --L.. 31. -^^i 32. ^.

c + ^ (^Y-1 x + l + l

e \h/ X

33. 1 34. 35. a + ^

1 .1 d

x-\ 4h cH

x-S 9

a , h ^ db 1,1 a2 + 62 ^

o«a + 6 a-6 a2-62 i + x 1-x „„ a a2-62

OO. • Of. • So. '

1 1 1111 a3 + 63

(a + &)2 {a-hY 1 - « 1 + ic a h

39 • a;-3y ^ x + 3y ^^ _J^ a; + 2

■ x2-2a;y-15?/2 " x2-8xi/ + 15y2' ' 3(x + l) 3(-4-3x + x2)

^^' 1 ri'+ 2y. ;•

X y + z
-Q 2a-36 + 4 3a-46 + 5 a-1

6 8 12

43 l ^"^ -^y^ x2 - y2 \ / x + y X - y \
■ \x2 - 2/2 a;2 4. 2/2/ • \^a; _ y x^yj'

X- 1 y- 1 z -1

-^ 3xy2 x__ y g

' yz -{• zx — XV 11 1

X y «
45 / 2x + y 2y-x x2 \ x2 + y2

\ X + y X - y x2 - y2/ ■ x2 - y2
-^a — 36 4a — 6 5a + 3c a2 — 6c 2a

40. 1 1 •

6a 26 9c 2ac 6

. „ 6cd! cda

47. 1 ^

(a - 6) (a - c) (a - d) (6 - c) (6 -d^(b-a)

dab ^

(c-6)(c-a)(c-d)
a6c
"^ {d -a){d- 6) (d - c) '

48 1 a-26 8 3a-4c2 9 5c2_--_66
' 6a 3a6 46 8ac2 8c2 126c2

' X-1 X + 1 (X - 1)2 (X + 1)2 X2 - 1 (X2 - 1)2

CHAPTER IV

EQUATIONS

52. Introduction. An equation is a statement of equality
between two expressions.

We assume the following

Axiom. If equals he added tOy subtracted frorriy multiplied hy^
or divided hy equals, the results are equal.

As always, we exclude division by zero. In dividing an equation by an alge-
braic expression one must always note for what values of the letters the divisor
vanishes and exclude those values from the discussion.

53. Identities and equations of condition. Equations are of
two kinds :

First. Equations that may be reduced to the equation 1 = 1 by
performing the indicated operations are called identities.

Thus 2 = 2,

a -6= (3a -26) -(2a -6)

are equations of this type. In identities the sign = is often replaced by =. It
should be noted that identities are true whatever numerical values the letters
may have.

Second. Equations that cannot be reduced to the form 1 = 1,
but which are true only when some of the letters have particular
values, are called equations of condition or simply equations.

Thus z=2 cannot further be simplified, and is true only when x has the value
2. Also a; = 2 a is true only when x has the value 2 a or a has the value - • If in
this equation x is replaced by 2 a, the equation of condition reduces to an identity.

The number or expression which on being substituted for a
letter in an equation reduces it to an identity is said to satisfy
the equation.

Thus the number 5 satisfies the equation x^ — 24 = 1. The number 3 satisfies
the equation (x — 3) (a; + 4) = 0.

32

EQUATIONS 33

The process of finding values that satisfy an equation is called
solving the equation. The development of methods for the solu-
tions of the various forms of equations is the most important
question that algebra considers.

In an equation in which there are two letters it may be possible
to find a value which substituted for either will satisfy the equa-
tion. Thus the equation x — 2 a = Ois satisfied if x is replaced by

2 -a, or if a is replaced by -• In the former case • we have solved

for £c, that is, have found a value that substituted for x satisfies
the equation. In the latter case we have solved for a. In any
equation it is necessary to know which letter we seek to replace
by a value that will satisfy the equation, that is, with respect to
which letter we shall solve the equation.

The letter with respect to which we solve an equation is called
the variable.

Values which substituted for the variable satisfy the equation
are called roots or solutions of the equation.

When only one letter, i.e. the variable, occurs in an equation, the root is a num-
ber. When letters other than the variable occur, the root is expressed in terms
of those letters.

54. Linear equations in one variable. An equation in which
the variable occurs only to the first degree is called a linear equa-
tion. To solve a linear equation in one variable we apply the
following

EuLE. Apply the axiom (§ 52) ^o obtain an equation in which
the variable is alone on the left-hand side of the equation.

The right-hand side is the desired solution.

To test the accuracy of the work substitute the solution in the
original equation and reduce to the identity 1=1.

Since the result of adding two numbers is a definite number, and the same is
true for the other operations used in finding the solution of a linear equation, it
appears that every linear equation in one variable has one and only one root.

When both sides of an equation have a common denominator, the numerators
are equal to each other. This appears from multiplying both sides of the equation
by the common denominator and then canceling it from both fractions.

34 ALGEBRA TO QUADRATICS

EXERCISES

Solve :

- 4«-2 , 6a; Sx ^

^•-^ + T = T + '-

Solution : Transpose the term involving x,

4a;-2 5x 3x ^

... . ^. 32x-16 + 25x-30a; ^

Add fractions, = 5.

40

Clear of fractions and simplify, 97 x — 216

x = 8.
rt a(d^ + x^) , ax

dx d

d^ 4- x^ X

Solution : Divide by a, = c + -.

dx d

Transpose the term involving x,

d* + x2 X

dx d

= c.

Add fractions, = c.

dx

Clear of fractions and simplify, , _

d

X = -'

c

3. {a-l)x = b-x. 4. {a-x){l-x) = z^-l.

5. a(x-a2) = 6(x-62). 6. 2a; - fx = f « - 1 - |x + 2.

7. 8x - 7 + X = 9x - 3 - 4x. 8. .617 x - .617 = 12.34 - 1.234x.

9. 3(2x-.3) = .6 + 5(x-.l). 10. 7 - 5x + 10 + 8x - 7 + 3x = x.
11. (x-3)(x-4) = (x-6)(x-2). 12. f {x\[|(|x+5)-10] + 3}-8 = 0.
13. (H-6x)2+(2 + 8x)2 = (H-10x)2. 14. 6 = 3x+i(x + 3)-^(llx-37).

15. 2(x + 5) (X + 2) = (2x + 7)(x + 3).

16. (7ix - 2|) - [4| -UH- 5a;)] = 18^.

17. 6x - 7(11 - X) + 11 = 4x- 3(20 -X).

18. (a - 6) (X - c) + (a + 6) (« + c) = 2(&x + ad).

19. 2x - 3(6 + f X) + ^(4 - X) - ^(3x - 16) = 0.

20. 5x-2 = fx + fx + fx + T'ffa; + H« + i|a;.

21. (a - 6) (a - c + x) + (a + &) (a + c - X) = 2 a*.

22. 12.9x - 1.46X - 3.29 - .99x - llx + .32 = 0.

23. 6.7x - 2^7.8 - 9.3x) = 5.38 - 4|(.28 + 3.6x).

EQUATIONS S6

24.3-^ = ^-. 25. -^+l^c.

3 11 mx nx

3 X
26

x-J:^x-^^ 27 iil^^lll^?.

x-S x-^' ' f (6x + l) 3'

28 t(a^-4) ^1 29 2x2-3x + 5 ^2

'1(3x4-5) 6* ■7x2-4x-2 7*

3o.» + l = 12 + l 31. i:i-%i = ^--l

x2x9 J+x4^+x4

„rt« + &aj_c + cZx ««25x — 2_52x — 5

a + b ~ c + d' ' 3"7x-3~ 7 3x- 7*

„- ox ex , /x , „- X + a & X - 6 a

34. — + — -^ — = h. 35. — = + -■ .

bag b a a b

36. 5^ + i, = x-l. 3y 3x-19^5x-25^3

a X — 13 X + 7

3 12 1 3 2

38. L4-L^ = -l--L. 39. ^—^ + ^-^i^ + '-:z^ = o.

3 12 12 1

2'^x 3'^x S'x"^

^^x-8x + 12 „, 18 ,-a(2x + l) Sox- 46 4

4U. 1- = z H • 41. — = — •

x + 2 x-8 x + 2 36 66 6

.„ ax 6x 2a6 (a + 6)2x

'4(6. i f- = <

6 a a + 6 ab

43. 8ix - - - 3|x- 4^x + 1 = 0.

6

.. 1 a + b 1 a — b

44. + = +

a + b x a — b x

45 ^ (^ - ^) _L ^ + 8 _ 3(5x + 16)
X- 7 X -4~ 5X-28 '

46.^

a 6

^^^ox-A_^

X b^c — X «c2 — X _

5X-.4 1.3 -3x _ 1.8-8X

2 1.2

^48. -^^ + -1^- + -^ = 2.
6x + 2 15x4-6 3x4-1

49.I^^2-i(. + 3) + 6 = ^-(^±?>.
3 6^ ' 2

36

50.
51.
52.
53.
54.
55.
56.
57.
58.
59.
60.
61.
62.
63.
64.
65.
66.
67.

5x-l 3x + 2 ic2-30x + 2

3(x + l) 2(x-l) 6x2-6

3x-2 . 7x-3 . x + 100

= 10.

x + 3 x + 2 x2 + 5x + 6
36(x-a) x-62 6(4a + cx) ^Q

5a 166 6a

16X-27 x + 3 _ 6 + 3x 4x-7

21 6 ~ 2 3 "

a{b-x) b{c-x) _ a + 6 , /& , a\

6x ex X \c &/

5x-6 9-lOx 3x-4 3-4x

10

4-2x

3

3-

6
1.5x

7
4x2

6x-3 X-.5 3(2x-l)

^ +

6(2x-5) 2(2x-6) 3(2x-6)

x« + i-3x»-i 3x«-i-x« x« „

4x 4 2

ax — be 6x — ac_cx — 62 x — a x

ab c^ be c a

3x

X — 2a X — 26 X — 2c

6 + c

a+c— 6 a+6— c a+6+c

2x« + 7x»-i 7x"-44x«-i 4X" + 27x'»-i

9 5X-14

3x + 3 /x + 1

18

a(x-3) 6(x-3) a2(x-l) 62(x-l) ^^
6 a 62 "^ a2 ~ •

(m + n)2x nx _ c 3nx

mb 6 m (a — 6) 6

m (a — 6) 6

4(13x-.6) 3(1.2-x)_9x + .2 5 + 7x

6 "^ 2 -~20~"^~~r~"^'^-

(^^±^(x-a) + ^^i:iA'(x-6) = 2a(2a + 6-x).
6 a

ax — 6 ex — d (6n + dm)x + (6p + dg) _ a e
mx—p nx — q (mx - p) (nx - g) ~" m n'

EQUATIONS 37

55. Solution of problems. The essential step in solving a
problem by algebra is the expression of the conditions of the
problem by algebraic symbols. This is, in fact, nothing else than
a translation of the problem from the English language into the
language of algebra. The translation should be made as close as
possible, clause by clause in most cases. In general the result
sought should be represented by the variable, which for that
reason is often called the unknown quantity.

Example. What number is it whose third part exceeds its fourth part by
sixteen ?

Solution : " What number is it " is translated by x. Thus we let x represent
the number sought. " Whose third part " is translated by - . " Exceeds its

XX

fourth part " is translated by , i.e. the third part less the fourth part

leaves something. " By sixteen " gives us the amount of the remainder. Thus
the translation of the problem into algebraic language is
Let X represent the number sought.

^-^ = 16.
3 4

This equation should be solved and checked by the methods already given.

PROBLEMS

1. What number is it whose third and fifth parts together make 88 ?

2. What number increased by 3 times itself and 5 times itself gives 99 ?

3. What is the number whose third, fourth, sixth, and eighth parts
together are 3 less than the number itself?

4. What number is it whose double is 7 more than its fourth part ?

5. In 10 years a young man will be 3| times as old as his brother is
now. The brother is 7| years old. How old is the young man ?

6. A father who is 53 years old is 3 years more than 12 1 times as old as
his son. How old is the son ?

7. If you can tell how many apples I have in my basket, you may have
4 more than ^, or, what is the same thing, 4 less than i of them. How many
have I?

8. If Mr. A received ^ more salary than at present, he would receive
\$2100. How much does he receive?

9. A boy spends ^ of his money in one store and \ of what remains in
another, and has 24 cents left. How much had he ?

38 ALGEBRA TO QUADRATICS

10. A man who is 3 months past his fifty-fifth birthday is 4^ times as old
as his son. How old is the son ?

11. In a school are four classes. In the first is ^ of all the pupils ; in the
second, ^ ; in the third, j- ; in the fourth, 37. How many pupils are in the
school ?

12. A merchant sold to successive customers ^, |, and ^ of the original
length of a piece of cloth. He had left 2 yards less than half. How long
was the piece ?

13. How may one divide 77 into two parts of which one is 2| times as
great as the other ?

14. The sum of two numbers is 73 and their difference is 15. What are
the numbers ?

15. A father is 4i times as old as his son. Father and son together
are 27 years younger than the grandfather, who is 71 years old. How old are
father and son ?

16. The sum of two numbers is 999. If one divides the first by 9 and
the second by 6, the sum of these quotients is 138. What are the numbers ?

17. The first of two numbers whose sum is a is b times the second.
What are the numbers ?

18. If the city of A had 14,400 more inhabitants, it would have 3 times
as many as the city of B. Both A and B have together 12,800 more than
the city of C, where there are 172,800 inhabitants. How many are in
A and B ?

19. Two men who are 26 miles apart walk toward each other at the
rates of 3| and 4 miles an hour respectively. After how long do they meet ?

20. A courier leaves a town riding at the rate of 6 miles an hour. Seven
hours later a second courier follows him at the rate of 10 miles an hour.
How soon is the first overtaken ?

21. A can copy 14 sheets of manuscript a day. When he had been work-
ing 6 days, B began, copying 18 sheets daily. How many sheets had each
written when B had finished as many as A ?

22. The pendulum of a clock swings 387 times in 5 minutes, while that
of a second clock swings 341 times in 3 minutes. After how long will the
second have swung 1632 times more than the first?

23. The difference in the squares of two numbers is 221. Their sum is
17. What are the numbers ?

24. If a book had 236 more pages it would have as many over 400 pages
as it now lacks of that number. How many pages has the book ?

25. A man is now 63 years old and his son 21. When was the father
19 times as old as his son ?

EQUATIONS 39

' 26. If 7 oranges cost as much less than 50 cents as 13 do more than
50 cents, how much do they cost apiece?

27.. The numerator of a fraction is 6 less than the denominator. Dimin-
ish both numerator and denominator by 1 and the fraction equals |. Find
the fraction.

28. The sum of three numbers is 100. The first and second are respec-
tively 9 and 7 greater than the third. What are the numbers ?

29. Out of 19 people there were f as many children as women, and 1^
times as many men as women. How many were there of each ?

30. A boy has twice as many brothers as sisters. His sister has 5 times
as many brothers as sisters. How many sons and daughters were there ?

ySl. A dealer has 5000 gallons of alcohol which is 85% pure. . He
/wishes to add water so that it will be 75% pure. How much water must

V 32. How much water must be added to 5 quarts of acid which is 10% full
strength to make the mixture 8|% full strength ?

.^33. A merchant estimated that his supply of coffee would last 12 weeks.
He sold on the average 18 pounds a week more than he expected, and it
lasted him 10 weeks. How much did he have ?

34. At what time between 3 and 4 o'clock are the hands of a clock point-
ing in the same direction ?

35. At what time between 11 and 12 o'clock is the minute hand at right
angles to the hour hand ?

^^6. A merchant bought cloth for |2 a yard, which he was obliged to sell
for \$1.75 a yard. Since the piece contained 3 yards more than he expected,
he lost only 2%. How many yards actually in the piece ?

37. A man has three casks. If he fills the second out of the first, the
latter is still f full. If he fills the third out of the second, the latter is still
I full. The second and third together hold 100 quarts less than the first.
How much does each hold ?

38. A crew that can cover 4 miles in 20 minutes if the water is still, can
row a mile downstream in f the time that it can row the mile upstream.
How rapid is the stream ?

39. A cask is emptied by three taps, the first of which could empty it
in 20 minutes, the second in 30 minutes, the third in 35 minutes. How long
is required for all three to empty the cask ?

40. A can dig a trench in f the time that B can ; B can dig it in f the
time that C can ; and A and C can dig it in 8 days. How long is required
by all working together ?

40 ALGEBRA TO QUADRATICS

56. Linear equations in two variables. A simple equation in
one variable has one and only one solution, as we have abeady
seen (p. 33). On the other hand, an equation of the first degree
in two variables has many solutions.

For example, Sx + 7y = l

is satisfied by innumerable pairs of numbers which may be sub-
stituted for X and y. For, transposing the term in y, we get

X- 3 ,

from which it appears that when y has any particular numerical
value the equation becomes a linear equation in x alone, and
hence has a solution. Thus, when y = 1, ic = — 2, and this pair
of values is a solution of the equation. Similarly, cc = — 9, y = 4
also satisfy the equation.

57. Solution of a pair of equations. If in solving the equation
just considered, the values of x and y that one may use are no
longer unrestricted in range, but must also satisfy a second linear
equation, we get usually only a single pair of solutions. Thus
if we seek a solution, that is, a pair of values of x and y satisfying

Sx-\-7y = l,

such that also

x + y = -ly

we find that the pair of values x = — 2, y = 1 satisfy both equa-
tions. Any other solution of the first equation, as, for instance,
x=—9, y = 4, does not obey the condition imposed by the second.
Two equations which are not reducible to the same form are
called independent.

Thus 6a; -82/ -4 =

and 3 a; — 4 y = 2

are not independent, since the first is readily reduced to the second by transposing

and dividing by 2. They are, in fact, essentially the same equation. On the other

hand, . „

' X — 4y=2

and 3 X — 4 y = 2

are not reducible to the same form and are independent. Since dependent equa-
tions are identical except for the arrangement of terms and some constant factor,
all their solutions are common to each other.

= -%

EQUATIONS 41

This principle we may state as follows :

Two equations , i, , r.

ax + oy + c =

and a'% + &'?/ + C =

are dependent when and only when

£_ &^ _ c

Independent equations in more than one variable which have
a common solution are called simultaneous equations.

Two pairs of simultaneous equations which are satisfied by the
same pair (or pairs) of values of x and y and only these are called
equivalent.

Thus r3.+7a==l. ^^ ra==-

are equivalent pairs of equations.

58. Independent equations. We now prove the following

Theorem. If A=0 and B = represent two in
equations^ then the pairs of equations

^ = ^'(1) and {'''■^f^'!: (2)

are equivalent where a, b, c, and d are any numbers such that
ad — be is not equal to zero.

The letters A and B symbolize linear expressions in x and y.
Evidently any pair of values of x and y that makes both A = and
B = 0, i.e. satisfies (1), also makes aA-{-bB=0 and cA + dB = 0,
i.e. also satisfies (2). We must also show that any values of x
and y that satisfy (2) also satisfy (1).

For a certain pair of values of x and y let

aA-{-hB = 0, (3)

cA + dB= 0. (4)

Multiply (3) by c and (4) by a (§ 52).

Then acA + bcB = 0, (5)

acA + adB = 0. (6)

42 ALGEBRA TO QUADRATICS

Subtract (5) from (6) (§ 52),

Thus, by § 5, either ad — be = ot B = 0.

But ad — be is not zero, by hypothesis ; consequently -6 = 0.
Similarly we could show that ^ = 0.

Thus if we seek the solution of a pair of equations ^ = 0,
jB = 0, we may obtain by use of this theorem a pair of equiva-
lent equations whose solution is evident, and find immediately
the solution of the original equations.

59. Solution of a pair of simultaneous linear equations. The

foregoing theorem affords the following

KuLE. Multijply each of the equations by some number such
that the coejfficunts of one of the variables in the resulting pair
of equations are identical.

Subtract one equation from the other and solve the resulting
simple equation in one variable.

Find the value of the other variable by substituting the value
just found in one of the original equations.

Check the result by substituting the values found for both
variables in the other equation.

Example.

Solve 3x-l-7y = l, (1)

x + y=-\. (2)

Solution : Multiply (1) by 1 and (2) by 3,

3x + 7?/ = l,
3x + 32/ = -3

Subtract,

4y =

4

Substitute

in (2),

x + l =

1.
-1.

Check: Substitute

in (1),

« =

-2.

3.

(-

- 2) + 7 . 1 =

-6 + 7

60. Incompatible equations. Equations in more than one vari-
able that do not have any common solution are called incompatible.

EQUATIONS 43

Theokem. The equations

ax-\-by = c, ^ (1)

afx-\-b'i/ = c' '*(2)
are incompatible when and only when aV — ba^ — 0.

Apply the rule of § 59 to find the solution of these equations.
Multiply (1) by a' and (2) by a.

We obtain aa^x + a^by — ca\

aa'x -f- ab'y = ac'.
Subtract, (ab' — a'b) y = ac' — ca'.

If now ab' — a'b is not zero, we get a value of y ; but since under
our hypothesis ab' — a'b = 0, we can get no value for y since divi-
sion by zero is ruled out (§ 7). Thus no solution of (1) and (2)
exists.
Example.

Solve Sx + 1y = l, (1)

6x + 14y = l. (2)

Solution : Multiply (1) by 2,

Qx + Uy = 2
6x+ Uy= 1
Subtract, 0=1

which is absurd. Thus no solution exists.

61. Resum^. We observe that pairs of equations of the form
ax -\- by -{- c = Oj
a'x -f J'y + c' =
fall into three classes :

(a) Dependent equations, which have innumerable common
solutions. ^ ^ ^

Tlien a'^b'^7'' W

(b) Incompatible equations, which have no common solution.
Then ^^, _ ^,j ^ ^^ ^^^ ^^^ -^ ^^^ ^^^^^

(c) Simultaneous equations, which have one and only one pair
of solutions.

Then ab' - a'b ^ 0.

18

44 ALGEBRA TO QUADRATICS

EXERCISES

Solve and check the following :
. 2x + 6y = l, 2 4x-6y,= 8, , 6x + 8y

' Qx + 1y = S. ' |x-2/ = f. •x + |y = 3.

^ 7x-3y = 27, ^ 2x-|2/ = 4, ^ |y = ^x-l,

* 6x-6y = 0. ■3x-|2/=:0. '^-^yrzfx-l,

„ 5x-4y + l = 0, g 3x + 4y = 253, g 6x + 3y + 2 = 0,

1.7x-2.22/ + 7.9 = 0. 'y = 5x. ■3x + 2y + l=0.

j^Q x + my = a, j^j^ x + y = |(5a + 6),

' x — ny = b. ' X — y = |(a + 6 6).

-2 2x-3y = -5a, -, |x-i(y + l) = l,

• 3x-22/ = -5&. • i(a; + l) + |(y-l) = 9.

-- 3.5x + 2i2/=13 + 4fx-3.5y, -- 3x + 2 ?/ = 5a2 + a6 + 5&2^
' 2ix + .8y = 22^ + .7x-3i2/. " 3?/ + 2x = 6a2 - a6 + 662.

16.

3 8

x+r

16 _ 4
X 2/

= 3,
= 4.

Hint. Ketain fractions.

18.

X - c
y-C

a
6'

r

x-y =

a -

h.

n/\

x + 2y
2x-2/

+ 1
+ 1

= 2,

17.

1 1_6

- 4- - — -1
X y 6

111

X y Q

3x + l_4

19. 4-2y 3
X + y = 1.

- + ^ = c,

20. r"^"^; 21. "^ ^

^^-^ + ^ = 5. ■ ^ + ^ = ci.

X — 2/ + 3 «! 6i

5 7

22.

24.

x + 2y 2x + 2/

7 ^ 6
3x-2~6-2/*
.9x-.7y + 7.3 _
13X-152/ + 17"
1.2x-.2y + 8.9 ^ ^
13x-16y + 17
X y 1

.2.

a + 6 a — 6 a — 6

X _ y _ 1
a + 6 a — 6 a + 6

x+1 a+6+c

2.3

y + l~a-6 + c'

X — 1 a + 6 — c

y-1 a-6— c

25.

y _6

27.

a- a - c
a2 - X a^-y _

EQUATIONS 46

^^^ y = 4 - 3x + x2. * • (4a; - 7) (X - 3) = y.

„Q4Vx-3Vy = 6, g-xVa-y\/6 = a + 6,

3 Vx — 4 Vy = 1. , X -\-y = 2 Va.

22 xV2 + ?/V3 = 3V3, 32 4V^T7-5Vir^ = 7,

'xV3-y\/2 = 2V2. "3 VxTT - 7 Vy^ = 2.

«^ Vx Vy „> Vx - 3 Vy + 3

12, ^^4 9 ,

—= + —= = 1. , + , = 4.

Vx Vy Vx - 3 V 2/ + 3

. ,, , a + 6 + 1 x + ly + 2 2(x-y)

(a_&)[x + (a+&)y]=a-6+l. 3(x - 3)-4(y- 3) = 12(2y -x>

62. Solutions of problems involving two unknowns. Tlie same
principle of translation of the problem into algebraic symbols
should be followed here as in the solution of problems leading to
simple equations (p. 37).

PROBLEMS

1. The difference between two numbers is 3|. Their sum is 9|. What
are the numbers?

2. What are the numbers whose sum is a and whose difference is 6 ?

3. A man bought a pig and a cow for \$100. If he had given \$10 more for
the pig and \$20 less for the cow, they would have cost him equal amounts.
What did he pay for each ?

4. Two baskets contain apples. There are 51 more in the first basket than
in the second. But if there were 3 times as many in the first and 7 times as
many in the second, there would be only 6 more in the first than in the
second. How many apples are there in each basket?

5. A says to B, "Give me \$49 and we shall then have equal amounts."
B replied, "If you give me \$49, I shall have 3 times as much as you.
How much had each?

6. A man had a silver and a gold watch and two chains, the value of the
chains being \$9 and \$25. The gold watch and the better chain are together
twice and a half as valuable as the silver watch and cheaper chain. The
gold watch and cheaper chain are worth \$2 more than the silver watch and
the better chain. What is the value of each watch ?

46 ALGEBRA TO QUADRATICS

7. What fraction is changed into ^ when both numerator and denomi-
nator are diminished by 7, and into its reciprocal when the numerator is
increased by 12 and the. denominator decreased by 12 ?

8. A man bought 2 carriage horses and 5 work horses, paying in all
\$1200. If he had paid \$5 more for each work horse, a carriage horse
would have been only J more expensive than a work horse. How much
did each cost?

9. A man's money at interest yields him \$540 yearly. If he had received
\% more interest, he would have had \$60 more income. How much money
has he at interest ?

10. A man has two sums of money at interest, one at 4%, the other at 5%.
Together they yield \$750. If both yielded 1% more interest, he would have
\$165 more income. How large are the sums of money ?

11. A man has two sums of money at interest, the first at 4%, the second
at 3^%. The first yields as much in 21 months as the second does in 18
months. If he should receive \% less from the first and \% more from the
second, he would receive yearly \$7 more interest from both sums. What
are the sums at interest?

12. What values have a mark and a ruble in our money if 38 rubles are
worth 14 cents less than 75 marks, and if a dollar and a ruble together
make %\ marks ?

13. A chemist has two kinds of acid. He finds that 23 parts of one kind
mixed with 47 parts of the other give an acid of 84 1% strength and that
43 parts of the first with 17 parts of the second give an 80f% pure mixture.
What per cent pure are the two acids ?

14. Two cities are 30 miles apart. If A leaves one city 2 hours earlier
than B leaves the other, they meet 1\ hours after B starts. Had B started
2 hours earlier, they would have met 3 hours after he started. How many
miles per hour do they walk ?

15. The crown of Hiero of Syracuse, which was part gold and part silver,
weighed 20 pounds, and lost 1^ pounds when weighed in water. How much
gold and how much silver did it contain if 19;^ pounds of gold and 10^
pounds of silver each lose one pound in water?

16. Two numbers which are written with the same two digits differ by
36. If we add to the lesser the sum of its tens digit and 4 times its units
digit, we obtain 100. What are the numbers ?

17. A company of 14 persons, men and women, spend \$48. If each man
spends \$4 and each woman \$3, how many men and how many women are in
the company?

EQUATIONS 47

63. Solution of linear equations in several variables. This
process is performed as follows :

KuLE. Eliminate one variable from the equations taken in
pairs, thus giving a system of one less equation than at first
in one less variable.

Continue the process until the value of one variable is found.

The remaining variables may be found by substitution.

Special cases occur, as in the case of two variables, where an infinite number
of solutions or no solutions exist. Where no solution exists one is led to a self-
contradictory equation on application of the rule. See exercise 17, p. 48.

EXERCISES

Solve and check the following :

« + 2/ + 2 = 9,
1. a; + 2y + 42 = 16,
x + 3y + 9z = 23.

Solution : x + y -[■ z = Q x -{■ y ■}- z— ^

x + 2y -{-iz = 15 x + 3y-f 9z=:23

2/ + 32= 6 2y + 8z = U

y + 4z= 7

y + Sz = 6

y -\-4z = 7

2=1

y + 3 = 6.

y = 3.

a; + 3 + 1 = 9.

x = S.

Check : 5 + 9 + 9 = 23.

X + y = 37, x + y = xy,

2.x + 2 =25, 3. 2x + 2z=x2;,

y + z = 22. Sz + Sy = zy.

Hint. Divide the equations by xy, xz, yz respectively.

X + y 4- z = 17, X + y + z = 36,

4. X + z - y = 13, 5. 4x = 3y,

x + 2;-2y = 7. 2x = 3z.

L3x-1.9y=.l, 2x+2y + « = a,

6. 1.7y-l.lz = .2, 7. 2y-i-2z+x = 6,

2.9x-2.1z = .3. 2z+2x + y = c.

48 ALGEBRA TO QUADRATICS

x + 2y = 6,

8.

y + 22; =8,

z + 2u = ll,

u + 2x = Q.

y z

10.

i + l = 25,

X z

1 + 1 = 20.
X y

"y _oo

iy-Zx--'''

12.

. ^' -15,

9.

X + y = m,
y + z = a,
z + u = n,
u — x = b.

11.

xy _1

; + y~5'
xz 1

x + z 6

yz _ 1

y-{-z~ 1

y + 1

13.?^ = 4,
X-32 ' z + 1

yz ^^2. 2; + 3_l

4y-5z x + 1 2

14. 312/ = X + 2 + 12, 15. X + 2 = 2| 2/ - 14,

4^ 2 = X + y + 16. y + 2; = 3f X - 32.

x + 2y-z=4.6, x + 2y + 32 = 15,

y + 22;-x = 10.1, 17. 3x + 5y + 7z = 37,
2 + 2x-y = 6.7. 5x + By + llz = 59.

7x + 6y + 72 = 100, (x+2)(2y + l) = (2x+7)y,

x_2y + 2 = 0, 19. (x-2)(3z + l) = (x+3)(32-]
3x + y-2« = 0. (l+l)(«+2) = (y+3)(z+l).

CHAPTEE V
RATIO AND PROPORTION

64. Ratio. The ratio of one of two numbers to the other is the
result of dividing one of them by the other.

a
The ratio of a to 6 is denoted by a : 6 or by - •

The dividend in this implied division is called the antecedent,
the divisor is called the consequent.

65. Proportion. Four numbers, a, b, c, d, are in proportion when
the ratio of the first pair equals the ratio of the second pair.

This is denoted by a : 6 = c : d or by - = - •

The letters a and d are called the extremes, b and c the means,
of the proportion.

66. Theorems concerning proportion. If a, b, c, d are in pro-
portion, that is, if

a:b = G:doT- = -j (I)

b d ^ ^

then ad = be, (II)

b:a = d:c, ' (III)

a:c = b:d, (IV)

a -\- b : a = c -{- d : Cf (V)

a — b : a = c — d : Cj (VI)

a-{-b:a — b=:c-{-d:c — d: C^II)

Equation (III) is said to be derived from (I) by inversion.
Equation (IV) is said to be derived from (I) by alternation.
Equation (V) is said to be derived from (I) by composition.
Equation (VI) is said to be derived from (I) by division.
Equation (VII) is said to be derived from (I) by composition
and division.

49

50 ALGEBRA TO QUADRATICS

67. Theorem. If a numher of ratios are equal, the sum of any
number of antecedents is to any antecedent as the sum of the
corresponding consequents is to the corresponding consequent.

Let a\h — c\d = e\f= g '.h^

or

a c e g
'h~d~f~h'

To prove

a + c + e b-hdi-f
g h

If

a c e g
1~'d~f~h~'''

we have

a = hr,

c = dr,

g = hr.
Divide the sum of the first three equations by the last and we get
gj^c + e ^ b + d-\-f
g h

68. Mean proportion. The mean proportional between two num-
bers a and c is the number b, such that

a:b =^b : c.
By (II), § 66, we see that ac = b^.

EXERCISES

If a : 6 = c : d, prove that :

a2 . , c2

1. a + h: = c + d:

a + b

Solution : By (V), § 66,
Squaring, we get

a + b c + d

a -hb _c + d
a c

(g + b)^ ^ (c + d)^
a2 c2 '

a + b _ c + d
^^ a2 ~ c2 '

a + 6 c + d
a + b: - = c + d

a + b c-\- d

I

RATIO AND PROPORTION 61

2. a2 : 62 = c2 : d2. 3. a + b:c + d = a:c.

4. ma:mh = nc: nd. 5. a2 : c2 = a2 + 62 . c2 + d^.

6. a2 + 62 : _^ = c2 + d52 : _^. 7. Va2 + c^ : VPTd^ = a : 6.
a + 6 c + cZ

8. ma + n6 : ra + s6 = mc + n(i : re + sd.

9. a + h + c + d:a — b-\-c — d = a + b — c — d:a — h — c + d.

10. Find the mean proportional between a^ + c^ and 62 + ^2.

11. Find the mean proportional between a2 + 62 + c^ and 62 + c2 + d\

Solve the following f or x :

12. 20:96 = x: 57.
14. x — ax:Vx = Vx : x.

- „ Vx + 7 + Vx 4 + Vx

10. — — ^;;^ =

V^rp^ -Vx4-v^ a-6a6 ac

Hint. Use composition and division.

18. (?^^^ + „i)..^±^-„i = (a + bf:z.
\a — I a + 6

13.

8 a6 : X :

= 6c : 1| ac.

15.

\--/x

l-3Vx =

= 1

4.

17.

a + & fl

^2-62
= X :

a -

-6

CHAPTER VI

IRRATIONAL NUMBERS AND RADICALS

69. Existence of irrational numbers. We have seen that in
order to solve any linear equation or set of linear equations with
rational coefficients we need to make use only of the operations
of addition, subtraction, multiplication, and division. When,
however, we attempt to solve the equation of the second degree,
x^ = 2, we find that there is no rational number that satisfies it.

Assumption. A factor of one memher of an identity between
integers is also a factor of the other memher.

Thus let 2 . a = 6, where a and b are integers. Then since 2 is a factor of the
left-hand member, it must also be contained in 6.

Theorem. iVb rational number satisfies the equation x^ = ^.
Suppose the rational number - be a fraction reduced to i'
lowest terms which satisfies the equation. Then

or a2 = 2 b\ (1)

Thus, by the assumption, 2 is contained in a^j and hence in a.
Suppose a = 2 a'.

Then by (1) 4 a'^ = 2 b%

or 2 a'^ = h\

that is, 2 must also be contained in ft, which contradicts the

hypothesis that t is a fraction reduced to its lowest terms.

The fact that the equation x^ = 2 has no rational solution is
analogous to the geometrical fact that the hypotenuse of an
isosceles right triangle is incommensurable with a leg.

62

IRRATIONAL NUMBERS AND RADICALS 63

70. The practical necessity for irrational numbers. For tlie
practical purposes of the draughtsman, the surveyor, or the
machinist, the introduction of this irrational number is superflu-
ous, as no measuring rule can be made exact enough to distin-
guish between a length represented by a rational number and one
that cannot be so represented. As the draughtsman does not use
a mathematically perfect triangle, but one of rubber or wood, it
is impossible to see in the fact of geometrical incommensurability
just noted a practical demand from everyday life for the intro-
duction of the irrational number. In fact the irrational number
is a mathematical necessity, not a necessity for the laboratory or
draughting room, as are the fraction and the negative number.
We need irrational numbers because we cannot solve all quad-
ratic equations without them, and the practical utility of those
nimibers comes only through the immense gain in mathematical
power which they bring.

71. Extraction of square root of polynomials. This process,
from which a method of extracting the square root of numbers is
immediately deduced, may be performed as follows :

EuLE. Arrange the terms of the polynomial according to the
powers of some letter.

Extract the square root of the first term, write the result as
the first term of the root, and subtract its square from the given
polynomial.

Divide the first term of the remainder hy twice the root
already found, and add this quotient to the root and also to the
trial divisor, thus forming the complete divisor.

Multiply the complete divisor hy the last term of the root and
subtract the product from the last remainder.

If terms of the given polynomial still remain, find the next
term of the root hy dividing the first term of the remainder hy
twice the first term of the root, form the complete divisor, and
proceed as before until the desired number of terms of the root
have been found.

54 ALGEBRA TO QUADRATICS

EXERCISES

Extract the square root of the following :

1. a* - 2 a^x + 3 aH^ - 2 ax^ + x\

Solution : a* - 2 a^x + 3 a'^x'^ - 2 ax^ + x* \a^ — ax + x^

a*

2a2-ax| - 2 a^x + 3 aH^ - 2 ax^ + x*
- 2 g^x + a^x^
2a^-2ax + x^\ 2 a^^ - 2 ox^ + x*
2 g^x^ - 2 ax8 + x^

2. 1 + x. 3. 1-x.

4. 3x2 _ 2x + x* - 2x3 + 1. 5. x* - 6x3 + 13x2 - 12x + 4.

6. x* + 2/4 + 2x3?/ - 2x2/3 _ a;2?/2. 7. 9x* - 12x3 + 34x2 - 20x + 25.

8. 49g4-42g36+37g262_i2a63+464. 9. 2g6- 2gc - 2&c H g2 + 62 4. c2.

10. W*102 + v*U^ + U>4v2 + 2 W3u2|0 _|_ 2 t>8t«2w + 2 W?3m2u.

72. Extraction of square root of numbers. We have the
following

Rule. Separate the mcmher into periods of two figures each,
heginning o.t the decimal point Find the greatest number
whose sqyiare is contained in the left-hand period. This is the
first figure of the required root.

Subtract its square from the first period, and to the remainder
annex the next period of the number.

Divide this remainder, omitting the right-hand digit, by twice
the root already found, and annex the quotient to both root and
divisor, thus forming the complete divisor.

Multiply the complete divisor by the last digit of the root,
subtract the result from the dividend, and annex to the remainder
the next period for a new dividend.

Double the whole root now found for a new divisor and pro-
ceed as before until the desired number of digits in the root
have been found.

In applying this rule it often happens that the product of the complete divisor
and the last digit of the root is larger than the dividend. In such a case we must
diminish the last figure of the root by unity until we obtain a product which is
not greater than the dividend.

IRRATIONAL NUMBERS AND RADICALS

55

At any point in the process of extracting the square root of a number before
the exact square root is found, the square of the result already obtained is less
than the original number. If the last digit of the result be replaced by the next
higher one, the square of this number is greater than the original number.

There are always two values of the square root of any number. Thus Vi = + 2
or — 2, since (+ 2)2 = (— 2)2= 4. The positive root of any positive number or
expression is called the principal root. When no sign is written before the radical,
the principal root is assumed.

EXERCISES

Extract the square root of the following :
1. 2.0000.

Solution :

2'.00'00'00'|1.414
1
2.4|1.00
96

281 1 400
281

2.824 1 11900

-A

11296
604

2. 96481.

3. 56169.

4. 3.

5. 877969.

6. 2949.5761.

7. 5.

8. 257049.

9. .00070128.

10. 99.

11. 69.8896.

12. .0009979281.

13. 12.

14. 49533444.

15. 9820.611801.

16. 160.

73. Approximation of irrational numbers. In the preceding
process of extracting the square root of 2 we never can obtain a
number whose square is exactly 2, for we have seen that such a
number expressed as a rational (i.e. as .a decimal) fraction does not
exist. But as we proceed we get a number whose square differs
less and less from 2.

Thus 1.2 = 1, less than 2 by 1.

1.42 = 1.96, less than 2 by .04.
1.41^ = 1.9881, less than 2 by .0119.
1.4142 = 1.999396, less than 2 by .000604.

5Q ALGEBRA TO QUADRATICS

Though we cannot say that 1.414 is the square root of 2, we may
say that 1.414 is the square root of 2 correct to three decimal
places, meaning that

(1.414)2 < 2 < (1.415)2.

74. Sequences. The exact value of the square root of most
numbers, as, for instance, 2, 3, 5, cannot be found exactly in deci-
mal form and so are usually expressed symbolically. By means
of the process of extracting square root, however, we can find a
number whose square is as near the given number as we may desire.
We may, in fact, assert that the succession or sequence of numbers
obtained by the process of extracting the square root of a number
defines the square root of that number. Thus the sequence of
numbers (1, 1.4, 1.41, 1.414, • • •) defines the square root of 2.

75. Operations on irrational numbers. Just as we defined the
laws of operation on the fraction and negative nimibers (pp. 2-4),
we should now define the meaning of the sum, difference, prod-
uct, and quotient of the numbers defined by the seqilence of num-
bers obtained by the square-root process. To define and explain
completely the operations on irrational numbers is beyond the
scope of this chapter. It turns out, however, that the number
defined by a sequence is the limiting value of the rational num-
bers that constitute that sequence, that is, it is a value from which
every number in the sequence beyond a certain point differs by as
little as we please. We may, however, make the following state-
ment- regarding the multiplication of irrational nimibers : In
the sequence defining the square root of 2, namely, (1, 1.4, 1.41,
1.414, • • •) we saw that we could obtain a number very nearly
equal to 2 by multiplying 1.414 by itself. In general, we multi-
ply numbers defined by sequences by multiplying the elements of
these sequences; the new sequence^ consisting of the products, defines
the product of the original numbers.

Thus (1, 1.4, 1.41, 1.414, • • ) (1, 1.4, 1.41, 1.414, • • •)
= (1, 1.96, 1.9881, 1.999396, • • •).

The numbers in this sequence approach 2 as a limit, and hence
the sequence may be said to represent 2,

IRRATIONAL NUMBERS AND RADICALS 57

76. Notation. We denote the square root of a (where a repre-
sents any number or expression) symbolically by Va, and assert

or, more generally, ^ _ ,

Va • V^ = Va • b.

Similarly, Va -^ V^ = Va -^ b.

EXERCISES

1. Form five elements of a sequence defining VS.

2. Form five elements of a sequence defining V6.

3. Form five elements of a sequence defining V6.

4. Form, in accordance with the rule just given, four elements of the
sequence \/2 • VS. Compare the result with the elements obtained in Ex. 3.

5. Form similarly the first four elements of product V2 • V6 with the
first four elements obtained by extracting the square root of 10.

77. Other irrational numbers. The cube root and higher roots
of numbers could also be found by processes analogous to the
method employed in finding the square root, but as they are
almost never used practically, they will not be included here.
It should be kept in mind, however, that by these processes
sequences of numbers may be derived that define the various
roots of numbers precisely as the sequences derived in the pre-
ceding paragraphs define the square root of numbers.

The Tith root of any expression a is symbolized by Va. Here
n is sometimes called the index of the radical. The principle for
the multiplication and division of radicals with any integral
index is given by the following

Assumption. The product (or quotient) of the nth root of two
numbers is equal to the nth root of the product (or quotient) of
the members.

Symbolically expressed,

Va • -y/b = VoT^,

Va H- V^ = Va -hb.

68 ALGEBRA TO QUADRATICS

78. Reduction of a radical to its simplest form. A radical is
in its simplest form when the expression under the sign is integral
(§ 11) and contains no factor raised to a power which equals
the index of the radical ; in other words, when no factor can be
removed from under the radical sign and still leave an integral
expression. We may reduce a quadratic radical to its simplest
form by the following

Rule. If the expression under the radical sign is fractional^
multiply both numerator and denominator by some expression
that will make the denominator a perfect square.

Factor the expression under the radical into two factors^ one
of which is the greatest square factor that it contains.
■ Take the square root of the factor that is a perfect square, and
express the multiplication of the result by the remaining factor
under the radical sign.

If the radical is of the nth index, the denominator must be made a perfect nth
power, and any factor that is to be taken from under the radical sign must also be a
perfect nth power.

EXERCISES

Reduce to simplest form :

1. Vv.

Solution :

/T2 /12 . 5 /4 • 15
\5=\ 25 =\ 25 ~-

= |Vl6.
5

2. vi-

3. V32.

4. Vf.

5. V27.

6. VA.

7. V243.

8. V'250.

9. Vi + 4.

10. Vi-|.

11. 8V75.

12. tV2762.
15. V¥ + |.

13. lV80a;8y*.

14. VtV + ^V

16. v-V^ + ^V

"■ xS-

■ ^^ S

■'■ xlf ■

"M

-->/!•

«■f^/S■•

-#

-• ^^m■

- -Vi^-

IRRATIONAL NUMBERS AND RADICALS 69

26. Vx^ - 2 x2y + X2/2. 27. V6x8-20x2 + 20x.

28 M^-^«'^ + « 29 / 2a;«-I2x2:+l8x

„Q / 2 gs - 8 a2 + 8 g „- j a^ + a% -~ab'

' \8x-8x2 + 2x3' * \ 9(g-&.)

g62 _ 63

79. Addition and subtraction of radicals. Radicals that are of
the same index and have the same expression under the radical
sign are similar. Only' similar radicals can be united into one
term by addition and subtraction. We add radical expressions by
the following

Rule. Reduce the radicals to he added to their simplest form. ,
Add the coefficients of similar radicals and prefix this sum
as the coefficient of the corresponding radical in the result.

A rule precisely similar is followed in subtracting radical
expressions.

EXERCISES

Add the following :

1. V27, \/48, and V75.
Solution :

V27 = -

n/ 9 • 3 = 3 V3

V48=-

v/I6.3= 4V3

\/75=^

v/25.3= 6V3

Sum = 12 \/3

2. V3 + 2V3. 3. 8V7-3V7. 4. ay/x-hy/x.

5. g + 2Vg + 3\/a + 2 Vl6g - \/27g.

6. 3 V8 + 4 V^ ~ 5 V50 - 7 V72 + 6 V98.

7. 8Vg+5\^-7Vg + 4Vg-6Vx-3 Vg.

8. 7 V4x + 4 Vox + 3 ViSx - 5 V36x - 2 V80x.

9. Vg^^ + Vl6g - 106 + Vgx2 - 6x2 _ V9(g - 6).

10. 4 Vo^ - 3 y/¥x + 2 Vc^ + Vd2x - 2 V(6 + dfx.

11. 6 Vx + 3 V2x - 5 V3x - 2 V4x + Vl2x - \/l8x.

60 ALGEBRA TO QUADRATICS

80. Multiplication and division of radicals. For these pro-
cesses we have the following

EuLE. Follow the usual laivs of operation (§ 10), using also
the assumption of § 77.

Beduce each term of the result to its simplest form.

" The operations of this section are limited to the case where
the radicals are of the same index. Radicals of different indices
as Vs and V^ must first be reduced to the same index. See § 87.

EXERCISES

1. Multiply V2 - V3 by V2 - VS.
Solution : V2 - V3

V2-V8

2 _ V6 - Vl6 + V2i
= 2-V6-4 + 2V6

= - 2 + V6.

2. Divide ^_tZ_ by v^ + \/y.
Vxy
x-\-y

y/xy ( Vx2 - Vxy + Vy2) ( Vx + y/y) *
Solution : — — —— = — ^ ^

Vx + y/y y/xy ( Vx + Vy)

_ Vx2 Vxy Vy2
Vx2/ Vxy V^

<^--</!-

Carry out the indicated operations and simplify :
3. VlO . V5. 4. V^ . ■^.

5. V28 . V7. • 6. Vf ^ Vf-

7. (a -6x^1 8. VS-Vff.

9. ( V7 - V3) ( V3 - V2). 10. (- 1 + V3)^

11. (6 V3 + V6) (6 V2 - 2). 12. y/lTc • VTOc.

* Since as + 63= (a + ft)(a« - a6 + ft*), if o= V^, 6= Vy we have

IRRATIONAL NUMBERS AND RADICALS 61

13. (a + 6-V^)(Va+V&). 14. (8 + 3 Vs) (2 - VB).

15. VVx + Vy • V Vx - s/y, 16. V6 + 2 V6 . V 6 - 2 V5.

17. Vx + Vjc2 - 1 . Vx - Vx2 - L 18. (x2 + y2) ^ (a; ^ _,. y ^),

19. --V^. 20. (4Va-V3^)(V^ + 2V3x).

X

21. VS.Jp.
\4a

25. (^.^Ij. 36.V^.Vg

27, ^*-^

+ 66
y ' \x/ \ ox^ — to2

:fl^.(.t^-,%). 28. (5-l^).(l - 1).

Vxy \y ^1 VVx Vy/

29. (2 V6 - Vl2 - V2i + Vis) V2.

30. (3\/8 + Vl8+V60-2\/72)V2.

31. (5 v^ - 4 V32 + 3 V60 - 3 V54) V3.

32. ( V9^T6 + 3 A^)(V9x + 6 - 3 \^).

33. [(V7 4 V3 + Vl0)(V7 + V3-Vi0)7.

34. (2 V30 .. 3 V6 + 5 V3) ( V8 + V3 - Vs).

35. (2V^ + V8-Vl2)(^V30-fV3+V2).

37. Find the value of ^ V2i - V| + 2 Vs - V6- V3 4- V5 to three
decimal places.

81. Rationalization. The process of rendering the irrational
numerator (or denominator) of a fractional expression rational
without altering the value of the fraction is called the rationaliza-
tion of the numerator (or denominator) of the fraction.

This is usually accomplished by multiplying both numerator
and denominator of the given fraction by a properly chosen
radical expression called the rationalizing factor.

62 ALGEBRA TO QUADRATICS

The principles in accordance with which this rationalizing
factor is selected are the following:

Principle I. Since {a ■i-b)(a — b)=a^ — b% the rationalizing
factor of -sfx ± Vy is sjx hF Vy.

Principle II. Since {a^ — a'b-\-lF) (a + 5) = a' + h^, the ration-
alizing factor of -sfx -\- Vy is V^ — V^ + Vy^ and conversely.

Since (a^ -\- ah -\- If) (a — b)=a^ — b^, the rationalizing factor
of -y/x — -Vy is V^ + Vxy + Vy^, and conversely.

EXERCISES

^Nationalize the denominators of the following :
- Va + Vx
Va — Vx
Solution : By Principle I the factor which will render the denominator

rational is Va + Vx.

Va + Vx Va + Vx Va + Vx a + x + 2 Vox

Thus
2.

Va—Vx Va—Vx Va + Vx
1

2+V2 + V3
Soltttioii : This problem requires a twofold rationalization.

1 (2+V2)-V3

2 + V2 + V3~ [(2+ V2) + V3][(2 +\^) - V3]
^ 2+-v^-V3 ^ 2+V^-V3
~ (2 + V2y-S ~4 + 4V2 + 2-3
_ 2+V2-V3 _ (2 + v^-V3)(3-4\^

3 + 4v^ ~ (3 + 4V2)(3-4v^)
_ 6 + 3-v^-3V3-8\/2-8 + 4V6
"" 9 - 16 . 2

2 + 6v^+3V3-4V6

^2-V3
Solution :

^2-2 + ^6 + ^^ Vi + v'e + v^

^-.■^S ^-^3 ^2+^6+V3-« 2-3

= ^(^ + V6 + ^9).

IRRATIONAL NUMBERS AND RADICALS 63

5 7-V5 g V3+V2

2+V3 3+V6 V3-V2

8 /- 3,

V3+V2 V2-V4 V2+3Vi

10. ^/^«±^. 11. I^±^^.

\a-y/x \a-Va2-l

12 2V6 ^2 1+3V2-2V3

* V2+V3+V6 ' V6_}.V3+V2

-- 2\/T6 -g Va + ic + Va — x

V3 + V5 + 2 \/2 Va + x - Va - x

16 2 j^ V6-V5-V3 + V2

Va + 1 + Va- 1 V6 + V5 - V3 - V2

18. Show that ^^ ~ "^ = - .10 • ■ -.

V2-fV3

19. Show that ^^^-^^ = 17.48 ....

V8-

-V7

20

V(l + a)(H-6)-

-V(i

-a)(l

-&)

21.

V(l + a)(l + 6) + V(l
Show that ^^+^^ -

-a)(l

-h)

V6_

= .168.

V3 + V5 + V6 - V5

82. Solution of equations involving radicals. We prove the
important

Theorem. WTien an equation in x is multiplied hy an expres-
sion in Xy the resulting equation has, in general, solutions which
the first one did not possess.

Let ^ =

represent an equation contaming x which is satisfied by the
values X = a, by ■ - n. Let 5 be an expression which vanishes
when X = ay p, • ' V. Then the expression

is satisfied not only when x = a, bj • - j n, but also when
x = a,l3,-';v. .

64 ALGEBRA TO QUADRATICS

Example. The equation x — 2 =

has X = 2 for its only solution, while the equation ,

(X - 2) (X - 3) =
has in addition the solution x = 3.

If in the course of a problem it is necessary to multiply an
equation by any expression involving the variable, the solutions
of the resulting equation must be substituted in the first one to
ascertain if any solutions have been introduced which did not
satisfy the original equation. Solutions which have been intro-
duced in the process of solving an equation, but which do not
satisfy the original equation, are called extraneous solutions.

It may be shown in a similar way that raising the equation in Xj

A =B,
to any power introduces extraneous solutions.

EXERCISES

1. Solve Vaj + 19 + Vx + 10 = 9.

Solution : '^ Vx + 19 = 9 - Vx + 10.

X + 19 = 81 + X + 10- 18 V»Tl0.
-72=- 18 Vx + 10.
4 = Vx + 10.
16 = x + 10.
x = 6."
Check : V6 + 19 + V6 + 10 =5 + 4 = 9.

2. Solve Va; + 19 - Vx + 10 = - 9.

Vx + 19 = - (9 - VxTlO).
Simplifying, we get x = 6.

Check : V6 + 19 - V6 + 10 =+6-4 = 1 9^-9.

Thus our result satisfies only the equation which was introduced in the
course of solving the problem, and is extraneous. The Original equation
has no solution.

Solve and check, noting all extraneous solutions :

3. VSx -1=5. • 4. Vi« -8 = 2.

5. V2x + V3x = 1. 6. Vx + V3x = 2.

7. V5x-7=v'4x + 3. 8. 5Vx-7 = 3v^-l.

IRRATIONAL NUMBERS AND RADICALS

65

9. Vx + 1 + Vx + 2 = 3.
11. 2\^-V2x = 2+v^.
13. V37 - 7 V6x^=^ = 4

10. Vl3 4-4v^^^

6.

12. Vx + 4 + \^+T=L

14. Vx2 - 7 x + 19 = v^^^.

15. 7 + Va;2-lla; + 4 = jc.

17. a; — Vax (1 + x) + 1 - x = 1.
19. |(7V5 + 6)-5=|(3v^-l).
I 4

16. 8 + V(x - 10) (X - 6) = X.
18. V2 (X + 1) + V2x + 15 = 13.
20. 2\/3 + 3V2x = 3V2 + 2V3a

21.

6 4- Vx 8

23. V7 X + 2 =

25. V2X-1

-Vx

6x + 6
V7X4-2'

2 (x - 3)
V2x-10'

22.

24.

4.

5 + v^
5-Vx
a — Vto 26

3 Vox

a + V6x 2 6 + 3 Vox
6x+ 10

26. V9 X + 10 =

27.

VTT-

VT

1 + VI -X 1 - VIT^

29. X — ax : Vx = Vx : X.

30. 2v^^T2 + V^T2 = i^^±^.

V8x + 8

1 + 2V3X-6 11 + 2 V3x- 5

V4X+9

28. Va - X + V6 - X =

V6

31.

1 + 3V3X-6 11 + 5V3X-5

32. V9x + 7 + V4X + 1 = V25x + 14.

33. Vx + 15 + Vx - 24 - Vx - 13 = Vx.

34. Vx - 7 + Vx - 2 - Vx - 10 = VxT5.

35. (VS-7)(Vx-3) = (Vx-6)(Vx-5).

36. (a + \^) Vx:(6-Vx) Vx = a + 1:6-L

37. (4V5-7):(5V^-6) = (V^-7):(V^-6).

38. ( Va V6 — V6 Va) Vx = a V6 Vx — 6 VaVx.

CHAPTEE VII

THEORY OF INDICES

83. Negative exponents. We have already seen (§ 16) that

a^-a"* = «" + "' (1)

when n and m are positive integers. We now assume that this law
still holds when one or both of the numbers m and n are negative
or fractional.

If we let a-m = ,

a'"

then

since the law (1) holds when n and m are any integers. This
notation may be expressed verbally as follows :

Principle. A factor of numerator or denominator of a frac-
tion may be changed from the numerator to the denominator,
or vice versa, if the sign of its exponent he changed.

84. Fractional exponents. Since (p. 57) Va • Va = a, it is
natural to devise a notation for Va suggested by the law (1).

If we let Va = a*,

we have Va • Va = a* • a^ = a' "^ * = a^ = a.

Furthermore, if we let -y^ — an
it would be consistent with law (1) to write

1 11 11 s

(a")'' = a« . a» = a" " = «"'.
This notation we shall assume in general. Thus

66

THEORY OF INDICES 67

With the adoption of this notation we can attach a meaning
:o any real number with any rational number for its exponent.
This notation may be expressed verbally in the following

Principle. The numerator of a fractional exponent indicates
a power, the denominator a root.

85. Further assumptions. The operation of multiplication is
subject to the following laws of exponents :

I. Commutative law of rational exponents :

(^a'^y = a""- = a'--'' =(a'-y.

II. Associative law of rational exponents :

(yj = a'?'- -^ = ««"•« = (a'^y.

The laws of operation (§ 10) defined for integral values of the
symbols we also assume when the symbols are expressions with
rational exponents.

86. Theorem. a''lf= {ctby, where r is any rational number as _.

We raise both sides of the equation a'b'' = (aby to the g-th
power separately and show that the results are equal.

Since r = ->

l(abyY = [(abyj = {aby = aPbP.

p p p p p p p p

Also (or by = (a«55) = (a^^»«)(a«6«) • • • {a'^^)

q terms

p p

P P P P

q terms

g terms

=iJm'

== a^hP,

{ariry

= l(abyj.

ing the qih

root and taking the principal

root,

we

obtain

a'b"-

= (aby.

•

68 ALGEBRA TO QUADRATICS

EXERCISES
yl. Express in simplest form with positive exponents :

, , 36a-26-ic-5

(a)

9a26-2c-i

Solution : By Principle, § 83, ^^^"^^~^^"^
9a26-2c-i

_46-i&+2c-ic+3
~ a2-a2

^46c
~ a* '

(b) (c) — -

^ ^ (xi2/l)-i ' 35x-2?/6z-4*

3a-i6-2 6a2x-i gft^c Va6v^

4X-22/-4' 56-ic2' ^'^ v'^&-^aJ62c*'

,.. 5 11 / — I — \ — ; /., V 4x-«y-3 15a'263-m

^"•^ • ' 5 a- 4 6-"' 14x«2/«-3

• 2. Arrange in order of magnitude the following :

(a) Vi, 71, 7|.

Solution : We first ask. Is (f )i > (|)^ ?

Raise both numbers to the sixth power.

We obtain (|)3 and (f)2,

or If and f,

er 2if and 2|.

Thus Vl>7l-

Now compare (f)^ and (|)io

Raise both numbers to the fourth power.
We obtain . (|)2 and |,

or 1| and If.

Thus VI > Vl.

THEORY OF INDICES 69

Now compare (§)' and (|)^.

Raise both numbers to the twelfth power.
We obtain (f)* and (|)3,

or fl and \\^-,

or 6yV and 6f|.

Thus -s/i>\^h

The order of magnitude is then Vli VI? Vf •

(b) V^, VI, </h (c) 4^, V8, -Ws.

(d) Vli, VI- (e) V3, ^5, Vl5.

3. Perform the indicated operations,
(a) \/2 . V3.

Solution : ^2 • V3 = 2' • 3^ = 2^ 3^

= (22 . 38)i = V'4^ = v^lOS.

V3

V5 . ''V3.V2

87. Operations with radical polynomials. These operations
follow the rules for the same operations previously given, pro-
vided the assumptions and principles of §§ 83-86 are observed.

EXERCISES

1. Divide x^ — y^ by ^/x — y/y.

2. Extract the square root of 4 x - 12 x^ 2/3 + 9 2/5 + 32 x^ - 48 yi + 64.
3x + 3x-i-6

3. Simplify

xi- 3x2 + 3x-^-x-i

4. Divide J-Va^ -J-V^ by sJ->

a'2 52

5. Extract the square root of 1 2.

6. Multiply - 3x-5 + 2^ by ^ - ?^.

X* x^ 6-1

7. Extract the square root of

-ix2y-2- i^yx-i + ^y2x-2-^xy-^ + 25?.
49 2 Id 7 7

8. Multiply V^ - x» + x^ f- Vx^ -x+v^-lby Vx + 1.

CHAPTEE VIII

88. Definition. An equation that contains the second bnt no
higher power of the variable is called "a quadratic equation. The
most general form of the quadratic equation in one variable is

ax'^ + hx-{-c = 0, (1).

where we shall always assume a, b, and c to represent rational
numbers, and where a =^ 0. Every quadratic equation in x can
be brought to this form by transposing and simplifying.

89. Solution of quadratic equations. The solution of a quad-
ratic equation consists in finding its roots, that is, the numbers
(or expressions involving the coefficients in case the coefficients
are literal) which satisfy the equation.

The common method of solving a quadratic equation consists in
bringing the member of the equation that involves the variable
into the form of a perfect square, i.e. into the form

x^-\-2Ax + A\

For example, let us solve

Transpose 8, x^-^2x = S.

If now we add 1 to both sides of the equation, the left-hand
member will be a perfect square,

x^-\-2x-{-l = 9.
70

Express as a square, (x + 1)^ = 9.

Extract the square root, x -\-l =±S.

Transpose, ic = — 4 or 2.

Both — 4 and 2 satisfy the equation, as we see on substituting

them for x. Thus

(-4)^ + 2(-4)-8 = 0,

and 2^ + 2 • 2 - 8 = 0.

Consider now the general case.

Let us solve ax'^ -{- bx -\- c = 0.

Transpose c, ax^ -\- bx =— c.

b c
Divide by a, x^ -^ - x =

/ bV
Add ( 77— I to both members to make the left-hand member a

\2aJ

perfect square,

b b'' _ c b"" _ -4.ac-\-b^

,a 4a^ a 4a^ 4 a^

4 ac

/ by b^-4

Express as a square, ( x + — — I = — —^
Extract the square root.

b V^^ — 4 ac

x-\-w- = ±

2a 2a

2a

Transpose, x = ^^ * (1)

The roots are

-b-\- V^^ - 4 ac _ -b - -s/b^ -4.ac
Xi — „ f X2 — „

2 a 2a

That the equation can have no other roots appears from § 96.

* This expression for the roots, _ / —

X = »

2a

may be used as a formula for the solution of a quadratic equation.

Thus to solve the equation 2a;2-3a; — 6=0

we may substitute in the formula a = 2, 6 = - 3, c = — 6, and obtain

3±\^+48 3±V57

Thus Xi-.

4 4

3+V57 3-V57

72 QUADRATICS AND BEYOND

One should verify the fact that both a-iid

— satisfy (1) and are consequently roots of the

2a

equation. They are, in geueral, distinct from each other. For
particular values of the coefficients to be noted later (§ 98) the
roots may be equal or complex (i.e. of form a -\- /3 V— 1, where
a and yS are ordinary rational or irrational nimibers).

We may sum up the process of solving a quadratic equation in
the following

EuLE. Write the expression in the form aa^ -\- hx -\- c = 0.

Transpose the term not involving x to the right-hand side of
the equation.

Divide both sides of ^he equation by the coefficient of a^.

Add to both members the square of one half of the coefficient
of X, thus making the left-hand member a perfect square.

Rewrite the equation, expressing the left-hand member as the
square of a binomial and the right-hand member in its simplest
form.

Extract the square root of both members of the equation, not
omitting the ± sign in the right-hand member.

Transpose the constant term, leaving x alone on the left-hand
side of the equation. The two values obtained on the right-hand
side by taking the + and — signs separately are the roots sought.

Check by substituting the solutions in the original equation,
which should then reduce to an identity.

90. Pure quadratics. A quadratic equation in which the coeffi-
cient of the term in x is zero is often called a pure quadratic. Its
solution is found precisely as in the general case, excepting that
we do not need to complete the square. Thus let us solve

ax^ -f- c = 0.

Transpose c, ax^ =— c.

Divide by a, ic* = — -.

Extract the square root, ^ = ± _,'
The roots are Xi = + -v/— -? x^ = ~ a/

EXERCISES

Solve and check the following :
1. 3x2 -6x- 10 = 0.

Solution : Transpose 10, 3 x^ — 6 x = 10.

. Divide by 3, x2 - 2 x = J/-

Add the square of \ the coefficient of x, i.e. 1, to both sides,

x2_2x + l = -Lo + l = Y-
Express as a square, (x — 1)2 = -LS-.

Extract the square root, x — 1 = ± "V^-

x = i±V¥.

Check: 3(l±^'-6(l±^)-10 = 0.

3±6^| + |-6:F6Vir_io = 0.

/13

2. 8x2 + 2x -3 = 0.

Solution : Transpose 3, 8 x2 + 2 x = 3.

Divide by 8, x2 + i x = f .

Add the square of \ the coefficient of x, i.e. ^^j, to both sides,

^' + \^ + -h = \ + iz = il'

Express as a square.

(^ + \Y = f f-

Extract the square root

X + 1 = ± |.

^=±'J-\

= -lov\.

Check: 8. (1)2 + 2. 1-

3 = 1 + 1-3 = 0.

8(-f)^ + 2(-f)-

-3 = 11-1-3 = 1-1-3=1-1

3. x2-ax = 0.

4. x2 = 169.

5. x2-ix = i

6. fx2 = 560.

7. x2 + x-l = 0.

8. 19x2 = 5491.

9. 3x2-7x = 16.

10. x2 = .074529.

11. 3x2+ ll = 5x.

12. x2 - tl X = 1.

13. x2 + X - 66 = 0.

14. 20x2 + x = 12.

1 = 0.

74 QUADRATICS AND BEYOND

15. 7a;2 + 9x = 100. 16. 6x2 + 6x = 66.

17. 14x2 - 33 = 71x. 18. 5x2 + 13 = 14x.

19. x2 - 8x + 15 = 0. 20. 91 x2 - 2 X = 45.

21. x2 + 2 X - 63 = 0. 22. x2 - 6x + 16 = 0.

23. x2 - lOx + 32 = 0. 24. 6x2 + 26^ = 25 ix.

25. 6x2 - 13x + 6 = 0. 26. 15x2 + 527 = 178x.

27. 2x2 + 15.9 = 13.6 X. 28. (x - 1)2 = a(x2- 1).

29. a'^{b-x)^=h^{a-x)^. 30. 13x2-19 = 7x2 + 5.

31. ax2 - (a2 + l)x + a = 0. 32. (a - x)(x - 6) = - ab.

33. 14 x2 + 45. 5 X = - 36. 26. 34. a^ {a - x)2 = b'^{b- x)2.

35. (a-x)2+(x-6)2 = a2 + 62. 35. (a -x)(x-6) = (a -x)(c-x>

37.^' = ^. 38. 2x + l = 3.
b d X

33 15x^810^ 40. x2 + ? = 50.

2 3x 7

- - 2 X 1050 yio« + a^,^ + a^ 5

41. — = 4<c. 1 = -.

3 7x 6 + xa + x2

43 ^ + ^^ :^ 2x + l 44 x3- 10x2 + 1 ^ ^ _ g
■x + 3 x+6* 'x2-6x + 9

.^ ax + b mx — n ac ^^ 1^ 4

45. = 46. = 0.

6x + anx — m xx — 2x-3

47. -— ^— = -. 48. — - — + = X - 1.

49.

te2 — mx + n n 9 2 x — 3

x-2 _ 3 (8 - x) gQ (a-x)3 + (x-6)8 _ a^-b^

}x+ 14~ 28-x ' * "

51.'^— + ^-^ = 12. 52.

2x- 3 X- I

53 5 + X ' 8 - 3 X _ 2 X -^

' 3 — X X x-2

55. (^'=8(^-15. 56.5^

\x -b) \x-bj 9

57. a2 - x2 = (a - X) (6 + c - x).

(a-

-X)-

-(X-

-6)

a

+ &■

16-

■ X

2(x

-11)

x-4

4

X •

-6

12

2x-
X —

T*

3x-

X -

4-1
-3

5x

X

-14
-4

5x-

^V

3x-

-1_

?4

■X —

58. i= + ^ = ^.

59 2_^±^ + J?- = ^Ili + ^Z^.
18 x + 4 4 6

60. {X - a + 6) (X - a + c) = (a - 6)2 - x^.

91. Solution of quadratic equations by factoring. When the
left-hand member of an equation can be factored readily, this is
the most convenient method of solution. It also illustrates very
clearly the meaning and property of the roots of the equation.

Example. Solve x^ -}- 2 x — 15 = 0.

Factor the left-hand member, (x + 5) (x — 3) = 0.

The object in solving an equation is to find numbers that substituted for
the variable satisfy the equation. But since zero multiplied by any number
is zero (p. 3), any value of x which causes one factor of an expression to
vanish makes the whole expression vanish. If in this case x = 3, our equa-
tion in factored form becomes

(3 + 5) (3 - 3) = (3 + 5) • =
and is satisfied. If we let x = - 5, the other factor becomes zero, and the
equation reduces to the identity

(5_5)(6-3) = 0(5-3) = 0.

Thus the numbers 3 and — 6 are solutions of the equation.

92. Solution of an equation by factoring. We have imme-
diately the

EuLE. Transpose all the terms to the left-hand member of the
equation.

Factor that member into linear factors.

The values of the variable that make the factors vanish are
roots of the equation.

EXERCISES

Solve and check the following :

1. 6x2 + x = 15.

3. 6x2 -X- 6 = 0.

5. 13x2-38x = 3.

7. x2-40x + 111 = 0.

9. x2-18x- 208 = 0.
11. x2-3ax-4a2 = 0.
13. (x2 _ 1) + (X - 1)2 = 0.

15. (2x - l)(x -I- 2) -f (X - 1) (X - 2).= - 4.

16. (7x - l)(x -f 3) - (4x - 3)(x - 1) =24.

2.

6x2-f-7x = 3.

4.

5x2-17x+6 = 0.

6.

2x2 -6x- 26 = 0.

8.

13x2 -40x + 3 = 0.

10.

3x2 -26x + 36 = 0.

12.

(X - a)2 - (x - 6)2 = 0.

14.

(3x_6)2_(9x-M)2 = 0.

76 QUADRATICS AND BEYOND

17. (3x + 1) (X + 1) - (4a; + 3) (x - 1) = - 2.

18. (X - a)(4ax - 6) + (x - 6)(4ax - 6) = 0.

19. 3x-4Vx^^ = 2(x + 2).

Solution : Transpose, x — 4 = 4 Vx — 7.

Square, x2 - 8 x + 16 = 16 x - 112.

Transpose, x2 - 24 x -f 128 = 0.

Factor, (x - 16) (x - 8) = 0.

The roots are x = 16. x = 8.

Check : 3-16-4 Vl6 - 7 - 2 (16 + 2) = 48 - 12 - 32 - 4 = 0.
3-8-4 VS~^ - 2(8 + 2) = 24 - 4 - 20 = 0.

In the following examples, as always, the quadratic equation should be solved
by factoring when possible. Recourse to the longer but sure method of completing
the square is always available.

When an equation is cleared of fractions or squared in the process of bringing
it into quadratic form (1), §88, extraneous solutions may be introduced. The
results should be verified in every case and extraneous solutions rejected.

20. a + Va2 - x^ = x. 21. VxT~5 = x - 1.

22. 2x-^V2x-l = x + 2. 23. 1- 6x + V5(x + 4) = 0.

24. Vll-x + \/x-2 = 3. 25. V2X + 1-2 V2x + 3 = 1.

Hint. Square twice.

26. Va(x-6) + V6(x-a) = x.

27. VxTS + V2X-3 = 6. 28. 2 V3 + X - 4 VS^^ = V60.

29. Vl + ox - Vl - ax = X. 30. V5x- 1 - V8-2x = Vx^.

31. Vx + 7-V5x-2 = 3. 32. x - 10 = |(x - 1)-V2x-1.

33. Va2"^^ + V&2 + X = a + &. 34. V4 - x + V6 - x = V9 - 2;

X

X \b

35. (a2-62)(x2+l) = 2(a2 + 62)x. 36. x + 2 a V2(a2 + 62) - x = 3a2 + 62.

37. xV^:r2 + 2V^T2zzV^H^. 38. -J^-^ + \f-^ = 2-

gg \^a-\-x-\-Va-x _a .^ Va + Vx _ 2 Vx (x+a)2
Va + X - Va - X X Va-Vx y/a + Vx a{x-a)

Hint. Rationalize.

Va — X 4- Vx — 6 ja — x ^^ ^ , y/b-Va_ 1 Va—Vb

41. V'J^ + v^ ^ /«E^. 42.^+;
Va - X - Vx-6 \ X - 6

V6 Vx Va

^2 2a-(l + a2)a; ^ 26 + (l + 62)x

4

44. 2v^+4-3V2x-3

Vx + 4

45. (a - a;)2 + (& _ x)2 = f (a - cc) (6 - x).

46. a&x2 - (a + 6) (a6 + 1) x + (a6 + 1)2 = 0.

47. V2x-2+V3x + 7 = V2x+ll + V3x-8.

48. V{a + x) (X + 6) + V(a - x) (x - b) = 2 Vox.

49. 2 72 a + 6 + 2 X - Vl0a + 6-6x = Vl0a + 96-6x.

93. Quadratic form. Any equation is in quadratic form if it

may be written as a trinomial consisting of a constant term and
two terms involving the variable (or an expression which may be
considered as the variable), the exponent in one term being twice
that in the other. By the constant term is meant the term not
containing the variable.

Thus X -S V ^ + 13 = 0, a ;-^ + x"^ - 3 = , a2x-2» _ (a + 6) a;-« + 62 ^ 0,
x2 — 2x — S— Va;2 _ 2 x — 3 + 17 =^ are all in quadratic form. In the last the
whole expression x2 _ 2 x — 3 is taken as the variable.

It is usually convenient to replace by a single letter the lower
power of the variable or expression with respect to which the
equation is in quadratic form.

EXERCISES

Solve and check the following :

1. X - 8 v^ + 15 = 0. (1)

Solution : Let y/x = y.

Then x = y2.

Substituting, (1) becomes y2 _ 3^ _j. 15 = 0.
Factor, (y - 6) (y - 3) = 0.

The roots are y = 5, y = S.

Thus Vx = 6, Vx = 3,

or X = 25, X = 9.

Check: 25 -8-5 + 15 = 0; 9-8-3 + 15 = 0.

78 QUADRATICS AND BEYOND

2. x-i -

5x-§ + 4 =

= 0.

Solution :

Let

X-3 = y.

Then

X-t = y1.

Substituting, (1) becomes

2/2_ 5y + 4 = 0.

Factor,

(y - 4) (y - 1) = 0.

The roots are

2/ = 4, y = \.

Since

x-l-^- ' .
x3 ^'

we have '

VX2 ^2

Thus

4'

X2 =

1 1

= 64' "^-=^8'

and

1'

X2 =

:1; x=±Vl = ±l.

(1)

3. 2Vx2-2x-3 + x2 - 2x =
Solution : Add — 3 to both members and rearrange terms,

x2 _ 2x - 3 + 2 Vx2 - 2x - 3 - 3 = 0. (1)

Let Vx2 - 2 X - 3 = y.

Then x2-2x -3 = 2/2.

Substituting, (1) becomes ^2 ^ 2 y — 3 = 0.

. Factor, (y + 3) (y - 1) = 0.

The roots are y = —3, y = 1.

Hence y2 _ ^2 _ 2 x - 3 = 9,

or x2_2x + l = 13.

Extract the square root, x — 1 = ± "s/l3.

The roots are x = 1 ± VlS.

Also x2 - 2 X - 3 = 1,

or x2-2x + l = 5.

Extract the square root, x — 1 = ± V5.

The roots are x = 1 ± VB.

4. x8 - 1 = 0. 5. x8 - 8 = 0.

6. x8-l = 0. 7. x5 + 8xi = 9x.

8. X - 6 x-i = 1. Hint. Divide by Vx. This factor cor-

responds to the root z = 0.

9- 4x3 + 5x^-1 = 0. 10. 10x4-21 = x2.

11. 2x^-3x^-\-x = 0. 12. 3v^ + 6-l^=4.

13. ax^p + bxp + c = 0. 14. 3 ic* - 7 x2 - 6 = 0.

15. 4x6 - 14x3 + 6 = 0. 16. x* - 13x2 + 36 = 0.

17. x~ 12Vx + ll = 0. 18. x3 + 4x^= 16ixVx2.

19. ■\/x3-2Vx + x = 0. 20. 8X-6 + 999x-3= 125.

21. 7^x6 + 5xv^ = 66. 22. 2(V5 - 3)=* - 3 = v^.

23. (x2 - 10) (x2 - 3) = 78. 24. (x2 _ 1)2 + (a;2 + i) = 2.

25. ( v^ - 1)^ + v^ = ^z. 26. x^ + x^ = (28 + 2-8)x V^xl

27. (Vx- 3)(v'x - 4) = 12. 28. (X + a)^ + (x + 6)* = (a - b)K

29. (2x2-3)2-(x2 + 4)2 = 7. 30. 2x2 + 3 Vx2 - x + 1 = 2x + 3.

31. (2x + 3)' + (2x + 3)-^:^0. 32. 8(8x - 5)8 + 5(5 - 8x)« = 85.

33. x4-4(a+6)x2+16(a-6)2 = 0. 34. 4x2 + 12xVTTi = 27(1 + x).

35. (2x2-3x+l)2=22x2-33x + l. 36. (x2_ 5x+7)2 -(x-2)(x-3) = l.

40. ^^ = x-8.

37. x2 + 5 = 8x + 2 Vx2 - 8x + 40. 38. 2 V(x - 4)2 + 4(x - 4)"^ = 9.

x-4
2 + x
42. a = x2 + 6 + -.

44.1+^- , ^' , =3.

2 2(l + vT+^)'

.- 7 , 41 Vx 97 . 5 ._ 2x (x-l)2 .

45. X* H = h x5. 46. h ^^ = 4.

X ^^ (x-l)2 2x

47. (X - 1)2 + 4 (X - 1) - 5 = 0. Hint. I.ety = -^^: then- = ^^~^^'' '

^ ' ^ ' " (X — 1)2 y 2x

48. (x2 + 2)^+ ^ =4x2 + 8.
Vx2 + 2

94. Problems solvable by quadratic equations. The principle
of translating the problem into algebraic symbols, explained in
§ 55, should be observed here. The result should be verified in
every case. It may happen that the problem implies restrictions
that are not expressed in the equation to which the problem
leads. In this case some of the solutions of the equation may
not be consistent with the problem ; for instance, when the

80 QUADKATICS AND BEYOND

variable stands for a nuinber of men fractional solutions should
be rejected. If only such results are obtained, the problem is
seK-contradictory. Often negative solutions should be rejected,
as when the result indicates a negative number of digits in a
number. Imaginary or complex (p. 72) results in general mean
that the conditions of the problem cannot be realized.

PROBLEMS

1. The product of ^ and ^ of a certain number is 500. "What is the
number ?

2. There are two numbers one of which is less than 100 by as much as
the other exceeds it. Their product is 9831. What are the numbers ?

3. The sum of the square roots of two numbers is VtI. One of the
numbers is less than 37 by as much as the other exceeds it. What are the
numbers ?

4. A man sells oranges for -^^ as many cents apiece as he has oranges.
He sells out for |3. How many had he ?

5. If the perimeter of a square is 100 feet, how long is its diagonal ?

6. A man sells goods and makes as much per cent as ^ the number of
dollars in the buying price. He made \$246. What was the buying price ?

7. Two bodies A and B move on the sides of a right angle. A is now
123 feet from the vertex and is moving away from it at the rate of 239 feet
per second. B is 239 feet from the vertex and moves toward it at a rate of
123 feet per second. At what time (past or future) are they 850 feet apart ?

8. What is the number twice whose square exceeds itself by 190 ?

9. What numbers have a sum equal to 63 and a product equal to 612 ?

10. The sum of the squares of two numbers whose difference is 12 is
found to be 1130. What are the numbers ?

11. By what number must one increase each factor of 24 • 20 so that the
product shall be 540 greater ?

12. What numbers have a quotient 4 and a product 900 ?

13. Two numbers are in the ratio of 4 : 5. Increase each by 15 and the
difference of their squares is 999. What are the numbers ?

14. If 4^ is divided by a certain number, the same result is obtained as
if the number had been subtracted from 4^. What is the number?

15. Separate 900 into two parts such that the sum of their reciprocal
values is the reciprocal of 221.

16. The denominator of a fraction is greater by 4 than the numerator.
Decrease the numerator by 3 and increase the denominator by the same,
and the resulting fraction is half as great as the original one. What is the
original fraction ?

17. The numerator and denominator of a fraction are together equal
to 100. Increase the numerator by 18 and decrease the denominator by 16,
and the fraction is doubled. What is the fraction ?

18. A number consists of two digits whose sum is 10. Reverse the order
of the digits and multiply the resulting number by the origingll one, and
the result is 2944. What is the number?

19. The sum of two numbers is 200. The square root of one increased by
the other is 44. What are the numbers ?

20. The difference of two numbers is 10, and the difference of their cubes
is 20630. What are the numbers ?

21. Around a rectangular flower bed which is 3 yards by 4 yards there
extends a border of turf which is everywhere of equal breadth and whose
area is ten times the area of the bed. How wide is it ?

22. Two bicyclists travel toward each other, starting at the same time
from places 51 miles apart. One goes at the rate of 9 miles an hour. The
number of miles per hour gone by the other is greater by 5 than the number
of hours before they meet. How far does each travel before they meet ?

23. A printed page has 15 lines more than the average number of letters
per line. If the number of lines is increased by 15, the number of letters per
line must be decreased by 10 in order that the amount of matter on the two
pages may be the same. How many letters are there on the page ?

24. A merchant buys goods for a certain sum. The cost of handling them
was 5% of their cost price. He sells for §504, gaining as much per cent as
T^Q the cost price was in dollars. What was the cost price ?

25. A man had \$8000 at interest. He increased his capital by \$100 at the
end of each year, apart from his interest. At the beginning of the third year
he had \$8982. 80. What per cent interest did his money draw ?

26. Two men A and B can dig a trench in 20 days. It would take A 9
days longer to dig it alone than it would B. How long would it take B
alone ?

27. A cistern is emptied by two pipes in 6 hours. How long would it
take each pipe to do the work if the first can do it in 5 hours less time than
the second ?

28. A party procures lunch at a restaurant for \$16. If there had been
5 less in the party, each member would have paid 15 cents more without
affecting the amount of the entire bill. How many were in the party ?

82 QUADRATICS AND BEYOND

29. A party pays \$12 for accommodations. Had there been 4 more in the
party, and if each person had paid 25 cents less, the bill would have been
\$16. How many were in the party ?

30. A grocer sells his stock of butter for \$16. If he had had 6 pounds less
in stock, he would have been obliged to charge 10 cents more a pound to
realize the same amount. How many pounds had he in stock ?

31. A man buys lemons for \$2. If he had received for that money 50
more lemons, they would have cost him 2 cents less apiece. What was the
price of each lemon ?

32. It took a number of men as many days to dig a ditch as there were
men. If there had been 6 more men, the work would have been done in 8
days. How many men were there ?

95. Theorems regarding quadratic equations. In this and the
following sections we prove several theorems concerning quad-
ratic equations. Similar theorems are later proved in general for
equations of higher degree.

Theorem. If a is a root of the equation

ac(^-\-hx-\-c = 0, (1)

then X — a is a factor of its left-hand member , and conversely.

The fact that a is a root of the equation is equivalent to the
assertion of the truth of the identity

aa^ -\- ha ■\- G =i Oj

by definition of the root of an equation (§ 53).
Divide ax^ 4- Jx -f c by ic — a as follows :

X — a\ ax^ -\- bx -\- c \ax + (^ 4- ao^)

(b + aa) X -{- c

(b -f- aa) X — aa^ — ba

aa^ -{- ba -\- c

Since the remainder vanishes by hypothesis, ax^ + bx -{- c is
exactly divisible hy x — a.

Conversely, we have already seen (p. 3) that ii x — a is a,
factor of an equation, a is a root, since replacing x hy a would
make that factor vanish.

EXERCISES

Form equations of which the following are roots.

1. 2, 6.

Solution : Since 2 and 6 are roots, (x — 2) and (x — 6) are factors, and the
equation having these as factors is

(X - 2) (X - 6) = x2 - 8 X + 12 = 0.

2. 1, - 1. 3.-3,-4.
4. V2, - V2. 5. - 2, - 6.

6. - 4 V2, V32. 7. V27, - 3 VS.

8. 2 + V3, 2 - V3. 9. a + V6, a - V6.

96. Theorem. A quadratic equation has only two roots.

Given the equation ax^ + ?)X + c = with the roots a and /3, to prove that
the equation has no other root, as 7, distinct from a and /3.

Since a and j8 are roots of the equation, x — a and x — /3 are factors.
Thus our equation may be written in the form

a(x-a:)(x-/3) = 0. (1)

If now 7 is a root, it must satisfy (1), i.e.

a(7-a)(7-/3) = 0.

But in order that any product of numerical factors should vanish one of
the factors must vanish. Tlius either a = 0, or 7 — or = 0, or 7-/3 = 0.
But, by hypothesis, 7 5^ <x and 7 ^ /S, so the last two factors cannot vanish.
Thus a = 0. This would, however, reduce our equation to a linear equa-
tion, which is contrary to our hypothesis that the equation is quadratic.
Thus the assumption that we have a third distinct root leads to a contradiction.

CoKOLLARv I. If a quadratic equation is satisfied by more than two distinct
values of the variable, then each of the coefficients vanishes identically.

The above proof shows that the coefficient of x^ must vanish. In the same
way it can be shown that b = c = 0.

Corollary II. If two quadratic expressions have the same value for more
than two values of the variable, then their coefficients are identical.

Let ax2 + 6x + c = a'x^ + b'x + c'

for more than two values of x. Transpose, and we obtain
(a - a')x^ + (6 - b')x + c - c' = 0.

We have then a quadratic equation satisfied by more than two values
of X. Thus by Corollary I each of its coefficients must vanish. Thus a' = a,
6' = 6, c' = c.

84 QUADRATICS AND BEYOND

This theorem taken with § 95 is equivalent to the statement that a quad-
ratic equation can be factored in one and only one way. Thus if

ax"^ + 6x 4- c = a (x — or) (x — /3),
we cannot find other numbers 7 and 5 distinct from a and /3 such that

ax2 + 6x + c = a (x — 7) (x — 5),
for then the equation would have roots distinct from a and /3.

97. Theorem. If the equation

x2 + 6x + c = 0, (1)

where b and c are integers, has rational roots, those roots must be integers.

P
For suppose - to be a rational fraction reduced to its lowest terms and

a root of (1).

Then

^, + ^-^ + = 0.
q2 q

or

p2 + ipq + c^2 = 0,

which gives

p'^ = - q{bp-{- cq).

Thus some

factor of q must be contained in p (§ 69), which contradicts

the hypothesis

that the fraction - is already reduced to its lowest terms.

98. Nature of the roots of a quadratic equation.

The equation

ax^ -^ bx -\- c =

(1)

has as roots (§ 89) x^ = "^ ^ ^^, (2)

_ ^ -l_ V^'^ -

- 4c ac

2a

-b-^b^-

- 4: ao

2a ■ (^)

These expressions afford an immediate arithmetic means of
determining the nature of the roots of the given equation when
a J b, and c have, numerical values and a =^ 0. In fact an inspec-
tion of the value of 6^ — 4 ac is sufficient to determine the nature
of the roots of (1). 'mM

I. When h^ — Jf.ac is negative, the roots are iikhg%naty (§89).
II. When y^ — Ji^ac^Oy the roots are real and equal. In this

b
case x^=x^=-—-

III. When y^ — Ji^ac is positive, the roots are real and distinct.

IV. When ¥ — Jiac is positive and a perfect square, the roots
are real, distinct, and rational.

In case IV, if J^ — 4 ac = A,

_ — h + Va _ — ^, - Va

^' ~ 2 a ' ^"^ " 2 a '

The converses of these four cases are also true. For instance,
if the roots of (1) are imaginary, from (2) and (3) it is clear
that P — 4: ac must be negative.

The expression A = Z>^ — 4 ac is called the discriminant of the
equation ax^ -\- bx -^ c = 0.

EXERCISES

1. Determine the nature of the roots of the following equations without
solving.

(a) 3x2 _4x-l =0.

Solution : A = (- 4)2 - 4 . 3 . (- 1) = 16 + 12 = 28 and is then positive.
Thus by III the roots are real and distinct.

(b) 3x2 - 7x + 6 = 0. (c) 6x2 - X - 1 = 0.
(d) 3x2 + 4x + l = 0. (e) x2-4x + l = 0.
(f) 2x2 -6x-9=:0. (g) 2x2-4x-2 = 0.
(h) 4x2 + 12x + 9 = 0. (i) 2x2 + 6x - 4 = 0.

(j) 4x2- 28x + 49 = 0. (k) 4x2 + 12x + 5 = 0.

2. Determine real values of k so that the roots of the following equations
may be equal. Check the result.

(a) (2 + A;)x2 + 2A:x + l = 0.

Solution : Here 2 -\- k = a, 2fc = 6, l = c.

Thus A = 62-4ac = 4fc2_4.(fc + 2).l

= 4 ^2 _ 4 A: _ 8.
Since the roots of an equation are equal when and only when its dis-
criminant equals zero (§ 98, II), the required values of k make A = and

are the roots of

4^2 _ 4^ _ 8 = 0,

or fc2 - A: - 2 = 0.

Solve by factoring,

k^-k-2=:{k-2){k + l) = 0.
Thus the values of k are /c = 2, k = — 1.

Check : Substituting in the original equation for /c = 2, we get
4x2 + 4x + l = (2x + l)2,
and for A: = — 1 we get x2 — 2 x + 1 = (x — 1)2.

86 QUADRATICS AND BEYOND

(b) x2 + fcx + 16 = 0. (c) a;2 + 2a; + A;2 = 0.

(d) x2-2A:x + l = 0. (e) 3A;x2-4x-2 = 0.

(f) A;x2-3x + 4 = 0. (g) x^ -\- 4kx -\- k^ + 1 = 0.

(h) A:2x2 4- 3x - 2 = 0. (i) (A:2 + 3) x2 + fee - 4 = 0.

(j) 3/cx2 + A-x - 1 = 0. (k) x2 + (3A; + l)x + 1 = 0.

(1) x2 + 3x + A; - 1 = 0. (m) x2 + 9A;x + 6fc + ^ = 0.

(n) 4 A;2x2 + 4 A-x - 125 = 0. (o) 2 x2 - 4x - 2 A; + 3 == 0.

(p) (A: + l)x2 + A:x + A; + 2 = 0. (q) A;x2 + (4 A; + l)x + 4 A: - 3 = 0.

(r) 2(A:4- l)x2 + 3A:x + A;-1 = 0. (s) (A: - l)x2 + 5A:x + 6A; + 4 = 0.

(t) (2A; + 3)x2-7A;x+^^^ = 0. (u) (A;-l)x2+(2A; +l)x+ A: + 3 = 0.

CHAPTEE IX
GRAPHICAL REPRESENTATION

99. Representation of points on a line. Let us select on the
indefinite straight line AB sl certain point as a point of refer-
ence. Let us also select a certain line, the length of which for
the purpose in hand shall represent unity. Let us further agree
that positive numbers shall be represented on ^J5 by points to
the right of 0, whose distances from are measured by the given

-3 -2 -1 +1 +2+3 1

numbers, and negative numbers similarly by points to the left.
Then there are certainly on AB points which represent such num-
bers as 2, — 3, ^, — 1^, or, in fact, any rational numbers. Since
we can divide a line into any desired number of equal parts, we
are able to find by geometrical construction the point correspond-
ing to any rational number. Furthermore, by the principle that
the square of the hypotenuse of a right triangle equals the sum
of the squares of the other two sides, we can find the point
corresponding to any irrational number expressed by square-root
signs over rational numbers. More complicated irrational num-
bers cannot, however, in general be constructed by means of ruler
and compasses, but we assutne that to every real number there
corresponds a point on the line, and conversely , we assert that to
every point on the line corresponds a real number. This assump-
tion of a one-to-one correspondence between points and real num-
bers is the basis of the graphical representation of algebraic
equations.

This amounts to nothing more than the assertion that every real number,
rational or irrational, as, for instance, — 6, 2 + V3, V3, tt, represents a certain
distance from on AB, and conversely, that whatever point on the line we may
select, the distance from to that point may he expressed by a real number. .

87

88

'<Y

X axis

T^

100. Cartesian coordinates. We have seen, that when the
single letter x takes on real values all these values may be repre-
sented by points on a straight line.
When, however, we have two variables,
as X and y, which we wish to represent
simultaneously, we make use of the
plane. Just as we determined arbitra-
rily, on the line along which the single
variable was represented, an arbitrary
point for the point of reference and an
arbitrary length for the unit distance,
so now we select an indefinite line along which x shall be repre-
sented, and another perpendicular to it along which 7/ shall be
represented. The former we call the X axis ; the latter the Y axis.
The intersection of these axes we take as the point of reference
for each. This point is called the origin.

We select a unit of distance for x and a unit of distance for 1/
-v^hich may or may not be the same, according to the problem
under discussion. As before, we represent positive numbers on
the X axis to the right, and negative numbers to the left. Positive
values of y are represented above the X axis, and negative values
below it. The arrowhead on the
axes indicates the positive direc-
tion. Any pair of values of x
and y, written (x, ?/), may now
be represented by a point on
the plane which is x units from
the Y axis and y units from the
X axis. Thus if x = 0, y = 0,
written (0, 0), the point repre-
sented is the origin. The point
(3, 0), i.e. a; = 3, ?/ = 0, is found
by going three units of x to the
right, i.e. in the positive direction of x and no units up. The point
(4, 3), i.e. a = 4, 2/ = 3, is found by going four units of x to the
right and three units of y up. The point (— 3, 1) is found by

3,4)

YA

0(0,7)

U',3.

3.0)

{4, -2)

GRAPHICAL REPRESENTATION 89

going three units of x to the left and one unit of y up. The point
(— 3, — 4) is found by going three units of x to the left and four
units of y down. In fact, if we let both x and y take on every
possible pair of real values, we have a point of our plane corre-
sponding to each pair of values of (x, y). Conversely, to every
point of the plane correspond a pair of values of (x, y). These
values are called the coordinates of the point. The value of a?,
i.e. the distance of the point from the Y axis, is called its
abscissa; the value of y, i.e. the distance of the point from the
X axis, is called its ordinate. If the point (x, y) is conceived as a
moving point, and if no restriction is placed .

on the value of the coordinates so that they

^ ., T 'PIT ^>^<^ Quadrant ist Quadrant

take on every possible pair oi real values, (-,+; (-|-,+)

every point in the plane is reached by the ^

moving point (X, y). 3rd Quadrant

The X and Y axes divide the plane into (-'-)
four parts called quadrants, which are num-
bered as in the figure. The proper signs of the coordinates of
points in each of the quadrants are also indicated.

EXERCISES

The following exercises should be carefully worked on plotting paper,
which can be bought ruled for the purpose.

1. Plot the points (2, 3), (0, 4), (- 4, 0), (- 9, - 2), (2, - 4).

2. Plot with the aid of compasses the points (l, V'2), (V3, — V2),
(2+V3, 2-V3),(-V2, -V2).

3. Plot the square three of whose vertices are at (— 1, — 1), (— 1,+ 1),
(+ 1, — 1). What are the coordinates of the fourth vertex?

4. Plot the triangle whose vertices are (2, 1), (—6, — 2), (—4, 4).

5. Plot the two equilateral triangles two of whose vertices are (6, 1),
(—6, 1). Find coordinates of the remaining vertices.

6. If the values of the coordinates (x, y) of a moving point are restricted
so that both are positive and not equal to zero, where is the point still free
to move ?

7. If the coordinates {x, y) of a moving point are restricted so that con-
tinually y = 0, where is the point still free to move ?

90

8. What is the abscissa of any point on the T axis ?

9. The coordinates of a variable point are restricted so that its ordinate
is always 2. Where may the point move ?

10. If both ordinate and abscissa of a point vanish, can the point move ?
Where will it be ?

11. Plot the quadrilateral whose vertices are (0, 0), (— 6, — 3), (5, — 5),
(—1, — 8). What kind of a quadrilateral is it?

12. The coordinates of three vertices of a parallelogram are (— 1, — 1),
(6, 2), (—1, — 6). Find the coordinates of the fourth vertex.

13. The coordinates of two adjacent vertices of a square are ( — 1, — 2) and
(1, — 2). Find the coordinates of the remaining vertices (two solutions). Plot
the figures.

14. The coordinates of two adjacent vertices of a rectangle are (— 1, — 2),

(1, — 2). What restriction is imposed on the coordinates of remaining
vertices ?

15. The coordinates of the extremities of the bases of an isosceles triangle
are (1, 6), (1, — 2). Where may the vertex lie? What restriction is imposed
on the coordinates (x, y) of the vertex ?

101. The graph of an equation. The equation a? = 2 ?/ is satis-
fied by numberless pairs of values (x, ?/); for example, (2, 1),
(0, 0), (1, ^), (— 2, — 1) all satisfy the equation. There are, how-
ever, numberless pairs of values which do not satisfy the equation ;

for example, (1, 2), (2, - 1), (- 1, 1),
(0, — 1). The pairs of values which
satisfy the equation may be taken
as the coordinates of points in a
plane. The totality of such points
would thus in a sense represent the
equation, for it would serve to dis-
tinguish the points whose coordi-
nates do satisfy the equation from
those whose coordinates do not.
After finding a few pairs of values which satisfy the above equa-
tion we note that any point whose abscissa is twice the ordinate,
i.e. for which cc = 2 ?/, is a point whose coordinates satisfy the
equation. Any such point lies on the straight line through the
origin and the point (2, 1). We can then say that those points

I

GRAPHICAL REPRESENTATION 91

and only those which are on the straight line represented in the
figure have coordinates which satisfy the equation. This line is
the graphical representation or graph of the equation.

The equation of a line or a curve is satisfied by the coordinates
of every point on that line or curve.

Any point whose coordinates satisfy an equation is on the
graph of the equation.

102. Restriction to coordinates. Iri § 100 it was seen that a
moving point whose coordinates were unrestricted took on every
position in the plane. We now see that when the coordinates of
a point are restricted so as to satisfy a certain equation (as x = 2 3/),
the motion of the point is no longer free, but restrained to move
along a certain path. Thus, for instance, the equation a; = 4 means
that the path of the moving point is so restricted that its abscissa
is always 4. Its ordinate is still unrestricted and may have any
value. This shows that the plot of a? = 4 is a straight line four
units to the right of the Y axis and parallel to it, for the abscissa
of every point on that line is 4, and every point whose abscissa is
4 lies on that line.

EXERCISES

Determine on what line the moving point is restricted to move by the
following equations. Draw the graph.

1. x = 6. 2. x = 0. 3. ?/ = f.

4. X = y. h. y = 2. 6. 3 X = y.

7. 2x = y. 8. y = 0. 9. x = -3.

10. X = 32/. 11. 3y = - X. 12. X + y = 0.

13. 6x = ll. 14. 2/=-3. 15. 2x = -3?/.

16. x = -2y. 17. x = -l. 18. 2x-62/ = 0.

103. Plotting equations. Plotting an equation consists in find-
ing the line or curve the coordinates of whose points satisfy the
equation. Thus the process of § 101 was nothing else than plot-
ting the equation x = 2y. This may be done in some cases by
observing what restriction the equation imposes on the coordinates
of the moving point ; but more often we are obliged to form a

92

table of various solutions of the equation, and to form a curve
by joining the points corresponding to these solutions. This
gives us merely an approximate figure of the exact graph which
becomes more accurate as we find the coordinates of points closer
to each other on the line or curve.

EuLE. Wlien y is alone on one side of the equation, set x equal
to convenient integers and compute the corresponding values of y.

Arrange the results in tabular form. Take corresponding
values of x and y as coordinates and plot the various points.

Join adjacent points, making the entire plot a smooth curve.

When X is alone on one side of the equation integral values of y may be assumed
and the corresponding values of x computed.

Care should be taken to join the points in the proper order so that the resulting
curve pictures the variation of y when x increases continuously through the values
assumed for it. By adjacent points we mean points corresponding to adjacent
values of x.

Any scale of units along the X and Y axes that is convenient may be adopted.
The scales should be so chosen that the portion of the curve that shows considerable
curvature may be displayed in its relation to the axes and the origin.

When there is any question regarding the position of the curve between two
integral values of x, an intermediate fractional value of x may be substituted, the
corresponding value of y found, and thus an additional point obtained to fix the
position of the curve in the vicinity in question.

EXERCISES
Plot:

1. x2 -4a; + 3 = y-

Solution : In this equation if we set x = 0,
1, 2, 3, etc., we get 3, 0, — 1, 0, etc., as cor-
responding values of y. Thus the points
(0, 3), (1, 0), (2, - 1), (3, 0), etc., are on
the curve. These points are joined in order
by a smooth curve.

X

y

X

y

3

-1

8

1

-2

15

2

-1

3

4

3

6

8

6

16

\

W^

1/

1

i

\

/

\

/

\

/

'

/■

i

\

/

\

f

\

/

*x

—

—

—

-

GRAPHICAL REPRESENTATION

93

2. y = x^-lx + l.
4. y = x^ - Sx + 2.
6. y = x^ — 2x + 1.
8. 2/ = 2x2-6x + 7.
10. y = 2x^-6x-3.

3. y = x2 + 1.

5. y = x^ — 4x.

7. y = x'^ + 6x + 5,

9. y = 2x2^-3x + 4.

11. 2/ = x2-12x + ll.

104. Plotting equations after solution. When neither x nor y
is abeacly alone on one side of the equation, the equation should
be solved for y (or x) and the rule of the previous section applied.
It should be noted that when a root is extracted two values of
y may correspond to a single value of x.

EXERCISES

Plot:

1. 2x2+ 3y2 = 9.

Solution

32/2 = 9
2/2 = 3

2x2,

3/ = ± V3-fx2.
Assuming the various integral values for x, we obtain the following

table and plot:

X

y

±V3 = ±1.7

1

±VI = ±i-5

2

±Vi=± -67

+ 3

imaginary

X

y

-1
-2
-3

±VJ = ± -57
imaginary

N:

In this example, when x is greater than 3 or less than — S, y is imaginary.
Thus none of the curves is found outside a strip x = ±S.

To find exactly where the curve crosses the X axis, the equation may be
solved for x and the value of x corresponding to y = found. Thus

X = ± VF^I^'-

If 2/ = 0, X = ± Vl = 2-1. This point is included in the plot
2. X2/ = 4.

Solution :

4

94

Form table for integral values of x.

i

X

y

-1

-4

-2

-2

-3

-f

-4

-1

-6

-1

-8

-\

-12

-i

Since this table does not give us any
idea of the curves between -\- 1 and — 1,
v^e supplement the table by assuming
fractional values for x.

3. x^ = y\
5. xy = — 1.

7. x2 + ?/2 = 16.

9. a;2 + 2/2 = 25.

11. 2x2/ + 3x = 2.

13. xy + 2/2 = 10.

10 — 2/2

Hint, a; =

y

4. X2/ = 16.

6. X2/ = X + 1.

8. x2 - 2/2 = 9.

10. x2 + X = 12 2/.

12. x2 + 9 2/2 = 36.

14. X — 2/ + 2 xy = 0.

2/

Hint. x = — -•

1 + 22/

2/

X

6

-1

8

-1

12

-i

y

-8
12

15. 6x2+2x + 32/2 = 0. 16. x2 + 2x + 1 = y2 _ 3^.

105. Graph of the linear equation. The intimate relation
between the simplest equations and the simplest curves is given
in the following theorems.

Theorem I. The graph of the equation y = ax is the straight
line through the origin and the point (1, a), where a is any real
number.

The proof falls into two parts.

First. Any point on the line through the origin and the point
(1, a) has coordinates that satisfy the equation. Let P (Figure 1)
with coordinates (x', y') be on the line OA. By similar triangles

V

or

ax'

GRAPHICAL KEPRESENTATION

95

Thus the coordinates of any point on the line satisfy the equation.

C«'y)

Figure 1

Figure 2

Second. Any point whose coordinates satisfy the equation lies
on the line.

Let the coordinates (x\ y') of the point P (Figure 2) satisfy the
equation. Then we have

or

y = ax',

^, = a.

x'

Let the ordinate y' cut the line at B. Then by the first part
of the proof BC = ax',

EC

or

x'

a.

Thus

Hence P lies on the line.

v' EC
a=—, = — TJ or ti' = EC.
x' x' -^

Theorem II. The graph of any linear equation in two vari-
ables is a straight line.

The general linear equation

Ax-{-By-{-C = (1)

may be written in the form

y = ax-\-b, (2)

A C

where a = — — and b= — —f provided ^ ^ (§ 7). It E = 0,
B E

the equation Ax -\- C = may be put in the form

C

96

provided A =^ 0. This is evidently the equation of a straight
line parallel to the Y axis (§ 102). li B = and A = 0,-we have
no term left involving the variable. Thus the only case for

which the theorem demands proof is
when B ^ 0, and the equation may be
reduced to form (2). By Theorem I
we know that the graph oi y = ax is
a straight line. If, then, we add to
every ordinate y of the line y = ax the
constant h, the locus of the extremi-
ties of the lengthened ordinates will
lie in a straight line, as one can easily
prove by Geometry. But any point (cc, y) on the upper line is
such that its ordinate y is equal to the ordinate of the lower
line, i.e. ax, and in addition the constant h ; that is, y = ax -\- h.
Also, since the upper line is the locus of the extremities of the
lengthened ordinates, every point whose coordinates satisfy the
equation y = ax + h i^ on this upper line. Thus the equation (1)
has a straight line as its graph.

CoEOLLARY. Two lines vjhose equations are in the form

y = ax-\-h, (3)

yz=ax-\-V (4)

are 'parallel. ^

For the value of the ordinates of (3) corresponding to a given
abscissa, say x^^ is obtained from the ordinate of (4) corresponding
to the same abscissa by adding the constant h — h\ Thus each
point on (3) is always found h — V units above (below if h — b'
is negative) a point of (4). Thus the lines are parallel.

106. Method of plotting a line from its equation. Since the
equation y = ax-\-h\^ satisfied by the values (0, h), the graph cuts
the Y axis h units above (below if b is negative) the origin. Since
it is satisfied by the values (1, a -\-b), the graph passes through
the point reached by going one unit of x to the right of (0, b) and
a units up (down if a is negative). These two points determine
the line. We may then plot a linear equation by the following

GRAPHICAL REPRESENTATION

97

Rule. Reduce the given equation to the form
y — ax-\-h.

Plot the point (0, h) as one of the two points that determine
the line.

From this point go one unit of x to the right and a units
of y up {down if a is negative) to find a second point that lies
on the line.

Draw the line through these two points.

EXERCISES

Plot:

1. 6x + 2y-5 = 0.

Write in the form

y = ax + h,

and we have

y=_3x + f.

a =-3;

6 = f.

Plot the point (0, |).

From this point go one unit of x to the right and three
units of y down to find the second point, which helps
determine the line.

2. 6i

32/ + ll = 0.

11

Yk

J

rV

/

/

io.

n

/

/

/

/

f

i

/

'

3. x-y = 0.

5. X + y = 4.

7. 2 X - y = 4.

9. Sx-y = 0.

11. x-8y = 16.

13. x = 8(2-2/).

15. X - y - 1 = 0.

17. x + 2/ + l = 0.

19. 12x-3y = l.

4. x — y — 5 = 0.

6. 2x = 6(l-y).

8. 12x + 10y = 5.
10. 16x-10y = 4.
12. 2x + y + 3 = 0.
14. 2x-6y-l = 0.
16. 2x-2y-6 = 0.
18. 3x-6y-4 = 0.
20. 7x-8y-9 = 0.

107. Solution of linear equations, and the intersection of their
graphs. The process of solving a pair of independent linear
equations consists in finding a pair of numbers (x, y) which
satisfy them both. Though each equation alone is satisfied by-
countless pairs of values (x, y), we have seen that there is only one
pair that satisfy both equations. Since a pair of values which

98

satisfy an equation are the coordinates of a point on its graph, it
appears that the pair of values that satisfy simultaneously two
equations are the coordinates of the point common to the graphs
of the two equations, that is, the coordinates of the points of
intersection of the two lines.

EXERCISES

Find the solutions of the following equations algebraically. Verify the
results by plotting and noting the coor-
dinates of the point of intersection.

1 3^-

^' 3x-

-iy

+ 16:

= 0,

-y -

-7 =

Solution

3x-

'4.y

+ 16 =

3x-

y

- 7 =

Sy

-23 =

.

y =

¥•

Substituting

in (2),

3x =

7 + ¥

X =

(1)

(2)

¥•

To plot (1) and (2) we get the equa-
tions in the form y = ax + h and apply
the rule. Thus

2/ = fx + 4.

2/ = 3x-7.

S'^^'iM

/\l

^ \

i^'f)'!

y

"{0.4)' i

^^

/~

z

/

y

- i I

"7^

t X

. 7

r

K'.-i)

'-t

.

JCo.r)

2 2x + 3y = 6,
7x + y = 2.

5.

3x-2y = l,
3x + 2?/ = 5.

g 2x + y = 3,
• 8x-72/ = l.

11 x + y = -i,
' 4x-3y = 5.

14.

17.

X + y = 5,
4x-2y = 28.

x-y = -4,
2x + 6y = 16.

6.

9.
12.
15.
18.

X + y = 5,

4

x-3y = -7,

3x + y = l.

4x + y = ll.

3x-7y = 9,

7.

X + 2/ = - 7,

X + 2 y = 3.

2x-3y = 6.

2x-5y = 0,
x-y = S.

10.

X + 2 y = - 10,
2x-y = 0.

x-y = 1,

13.

x-y = l,

2x-8y = 3.

2x-42/ = -16.

X - 2/ = 2,

16.

6x-5y = 5,

4x- 52/ = 9.

2 X + 3 y = — 20.

3x + 2|/ = 9,
8x-2/ = 2.

19.

2x + 6y = -20, •
3x + y = 2.

GRAPHICAL REPRESENTATION 99

108. Graphs of dependent equations. We have defined (§ 57) dependent
equations as those that are reducible to the same form on multiplying or
dividing by a constant. Thus two dependent equations are reducible to the
same equation of the form y = ax + b. Hence dependent equations have as
their graphs the same straight line. We see now the geometrical meaning of
the statement that dependent equations have countless common solutions.
Since their graphs have not one but countless points in common, being the
same line, it is clear that the coordinates of these countless points will
satisfy both equations.

109. Incompatible equations. By our definition (§ 60) incompatible equa-
tions have no common solution. Since every pair of distinct lines have a
common point unless they are parallel, we can foresee the

Theorem. Incompatible equations have parallel lines as graphs.

Let the equations

ax -{- by -{- c = 0, (1)

a'x + b'y + c' = (2)

be incompatible. This is true (§ 60) when and only when

ab' - a'b = 0. (3)

(4)
(5)

This may be done if neither 6 nor ¥ equals zero. If both b and b' vanish,
the lines (1) and (2) are both parallel to the Y axis and hence to each other,
which was to be proved. But if only one of them vanishes, say & = 0, then
by (3) a = (§ 5), in which case (1) does not include either variable. Thus
we may assume that neither b nor 6' vanishes and that (4) and (5) may be
obtained from (1) and (2).

By (3) a _a/

b~V

Our equations (4) and (5) become

a c

a c'
y = --x_-,

which represent parallel lines, by the Corollary, p. 96.

This theorem completes the discussion of the graphical representation of
the possible classes (§ 61) of pairs of linear equations.

Let us then assume (3).

Write (1) and (2) in the form

y =

a c

—b'-b

y =

100 QUADRATICS AND BEYOND

EXERCISES

Plot and solve :

1.

8x + 2y = 3,

4x + y = 8.

3.

10x-5y =
2 X - y = 3.

15,

5.

x-1y = l,

4x-2Sy =

56.

7.

x-Sy = 2,

6x-lSy =

36.

2x + 6y = 1,
x-^Sy = 1.

2x-Sy = 6,
8x- 12y = 24.

12x-6y=18,
2x-y=l.

2x-S + y = 0,
4x -7 -\-2y =0.

110. Graph of the quadratic equation. Let

y = ax^ -{- bx -\- c, (1)

where as usual a, b, and c represent integers and a is positive.

If we let X take on various values, y will have corresponding

values and we may plot the equation as in § 103. A root of the

ax^-}-bx + c = (2)

is a number which substituted for x satisfies the equation, that
is, gives the value y = in (1). Thus the points on the graph
of (1) which represent the roots of the equation (2) are the
points for which y = 0, that is, where the curve crosses the
X axis. The numerical value of the roots is the measure of
the distance along the X axis from the origin to the points where
the curve cuts the axis. Since this distance is always a real
distance, only real roots are represented in this manner.

Theorem. If the graph of (1) has no point in common with
the X axiSf the equation (2) has imaginary roots, and conversely.

Every equation of form (2) has two roots either real or imagi-
nary (§ 89). If the graph of (1) has no point in common with
the X axis, there is no real value of x for which y = 0, i.e. no
real root of (2). The roots must then be imaginary.

Conversely, if (2) has only imaginary roots, there is no real
value of X which satisfies it, i.e. which makes y = in (1).
Thus the curve has no point in common with the X axis.

GRAPHICAL REPRESENTATION 1,01

This suggests the following universal

Principle. Non-intersection of graphs corresponds to imagi-
nary or infinite-valued solutions of equations.

111. Form of the graph of a quadratic equation. Consider the

equation ^ o rr ^

^ y ^2x^+1 x-\-2. ^ (1)

By substituting for x a very large positive or negative number,
say X =± 100, y is large positively. Thus for values of x far to
the right or left the curve lies far above the X axis. If we
assign to t/ a certain value, say 3/ = 2, we can find the correspond-
ing values of x by solving a quadratic equation. Thus in (1) let

or . 2x^ + 1 x = 0.

The roots are x^=— 3^, x^ = 0.

Hence the points (— 3^, 2) and (0, 2) are on the curve (§ 101).
That is, if we go up two units on the Y axis, the curve is to be
found three and one half units to the left and also again on the
Y axis. If in (1) we let y = — 4, the corresponding values of x
are very nearly equal to each other (— 1^ and — 2), which means
that the curve meets a line parallel to the X axis and four units
below it at points very near together. The question now arises,
Where is the bottom of the loop of the curve ? This lowest point
of the loop has as its value of y that number to which correspond
equal values of x. Hence we must determine for what value of
y the equation (1), that is, the equation

2x^-{-lx + {2-y)=0,

has equal roots. Comparing with the equation ax"^ + Jx + c =

or

Z -

= a, 7 =

--b,2-

-y=-c.

Thus the condition h

2-4ac

= becomes

49-4

•2(2-

-2/)=0,

y =

49-
8

16_

33

" 8

4i.

102

I as the corre-'

Substituting this value of y in (1), we get
spending value of x.

This gives a single
value of y for which
the values of x are
equal; hence the graph
of (1) is a single fes-
toon as in the figure.

If we take the gen-
eral equation

ax^ -{- bx -{- c = y,

we find precisely similarly that the bottom of the loop is at a
point whose ordinate is

P — Aac A

y = -

y

X

2

-4

-.3+ or -3.2 +
or - ^
- li or - 2
-If

4a

4a

Thus we see again that if the discriminant is negative the graph
is entirely above the X axis and both roots are imaginary (§§ 98,
110), since the ordinate of the lowest point of the loop is positive.
If the discriminant is positive, the graph cuts the X axis and both
roots are real.

The results of this section enable us to determine a value of
y from the coefficients which determine the lowest point of the
loop of the curve precisely, and hence to show beyond question
from the graph whether the equation ^as real or imaginary roots.

EXERCISES

Plot the following equations and determine by measurement the roots in
case they are real. Find in each case the lowest point on the loop.

1. x2 + X + 1 = y.

4. x2 + 7 aj + 6 = y.

7. x2-6x + l = y.

10. x2 + 2x-l = 2/.

13. 2x2-x-3 = y.

3. x2-6x + 10 = ?/.
6. 3x2-7x-6 = y.
9. 2x2-9x + 7 = y.

2. x2 - 4 X + 7 = y.

5. x2-6x + 9 = y.

8. x2-6x + 5 = y.

11. x2 -4x + 4 = y.

14. 3 x2 + 8 X + 5 = y.

16. What is the characteristic feature of the plot of an equation whose

roots are equal ?

12. 3:

4x-3 = y.

15. 4x2+12x+9 = y.

*

GRAPHICAL REPRESENTATION 103

112. The special quadratic adc^ + 6ic = O. When in the quad-
ratic equation

ax^ -\-hx^ G = 0, (1)

c = 0, we can always factor the equation into

ax^ -\- bx = X (ax -j- b) = 0,

(—:)=»■

or

Thus the roots are £Ci = 0, X2=

Conversely, if a; '= is a root, then (§ 95) ic — 0, or x, is a factor
and the equation can have no constant term.
This affords the

Theorem. A quadratic equation has a root eqtcal to zero
when and only when the constant term vanishes.

We show in a similar manner that both roots of the equation
(1) are zero when and only when i = c = 0.

EXERCISES

1. Prove the theorem just given by considering the expressions for the
roots in terms of the coefficients (§ 89).

2. For what real values of k do the following equations have one root
equal to zero ?

(a) x2 + 6x - A: + 1 = 0. (b) 2x2 - 3x + A;2 - 1 = 0.

(c) x2 + 6a; ^. ^2 ^. 1 ^ 0. (d) 2x2 - 4x + fc2 _ 3^ ^ 0.

(e) 2x2 + 2fcx-2A;2-4A;-2=0. (f) 6x2 - 4x + 2 A;2 + fc + 7 = 0.

3. What is the characteristic feature of the plot of an equation which has
one root equal to zero ?

4. For what real value of A: will both roots of the following equations vanish ?

(a) - + 3x - 1 = 0. (b) x2 + (fc2 + 3)x + fc - 3 = 0.

(c) x2 + (fc2 + l)x + l = 0. (d) x2+(fc-3)x + 2fc2-5fc-3 = 0.

(e) x2 + (A: + l)x + /c2 -1 = 0. (f) (fc-3)x2 + (fc2_9)x + A;2-4A; + 3=a

104 QUADRATICS AND BEYOND

113. The special quadratic ax^ + c = O. This equation may
be written in the form x^ -\- - = and factored * immediately into

(^-*-^R)(^-^R)=o>

which shows that the roots are equal numerically but have oppo«
site signs. The roots are

Xi =\/-^' ^2

xR

Since in the equation ax^ -\- c = y the variable x occurs only in
the term a:^, we get the same value of y for positive and negative
values of x. Hence the loop which forms the graph of the equa-
tion is symmetrical with respect to the Y axis.

114. Degeneration of the quadratic equation. The equation

ax^ -\-hx -\- c =■
has the roots

_ J 4_ V52 - 4 ac

Xi —

2a

Xo =

We wish to find the effect on the roots x^ and x^ when a
becomes very small. If we let a approach 0, then x^ approaches

an expression of the form -? which must always be avoided.

Rationalize the numerators and we get

4 ac 2 c

2a{- b - ■\/b^ - 4 ac) -b-\^b^-4: ac
4 ac 2 c

2a{-b + -^b^-4:ac) - b -{- ^b^ -4.ac
As a approaches 0, evidently h^ — ^ac approaches J^, Xi ap-
proaches — j> and x^j since its denominator becomes very small,

• When — Is positive this involves real factors. If — is negative the factors are
a a

imaginary (§ 152).

GRAPHICAL REPRESENTATION

105

increases without limit, that is, approaches infinity. Thus the
quadratic equation approaches a linear equation when a approaches
0, and one of its roots disappears since it has increased in value
beyond any finite limit. The loop-shaped graph of the quadratic
equation must then approach a straight line as a limit when a
approaches 0. This is made clear from the following figure, where
a has the successive values 1, \, y^^, ^V, 0.

In the figure the curves represent the following equations :

x-'-l-2 = y..

(I)

x" X
5-2-2 = ^-

(H)

10-2-2 = 2'-

(III)

X^ X ^

50-2-2 = ^-

(IV)

-f-2-

(V)

%%-A "■

Ml II 171

^^^J

7 1

tsX^

t J~

>^ \^v ^

t t h-

^§^^A

l f - ^

^SSv^

t ^ y^^^

^^>

^ ^'^,^

<

N^^

^^^>^

^^^-L*-

^-.K

In a similar manner we can show that when in the equation
Jx + c = 0, & approaches as a limit, the root of the linear equa-
tion becomes infinite.

EXERCISES

1. "What real values must k approach as a limit in order that one root of
each of the following equations may become infinite ?

(a) Arx^ + Gx + l = 0.

(c) (A;x - 1)2 - (X + 2)2 = (A: + a:)2.

(e) V2A:x-H-\/6F=^=V^TT.

X

(b) (^2 + 1)3524. a; + 1 = 0.
(d) A;2 + 4fc2a;2_(aj_i)2 4.2

(g)

1 fc + 1 X
1 X + 1 A;2

0.

(f ) Vx-fc + Vx + A; = yfkx + 1.
1~\ fc2

/ fcx-1
'a:x + 1

(i)

(^

1)^ ^ (fc

(A; + 1) k

(X + 1)2
(j) (A:2 - l)x2 + (A: - l)x + A:2 + 4A; - 5

106 QUADRATICS AND BEYOND

2. What real values must k and m approach as a limit in order that both
roots of the following may become infinite ?

(a) A:x2 + mx + 1 = 0.

(b) (2fc-m)x2 + A:x-2 = 0.

(c) 2 tec2 + (3m - 1 + h)x = 8x2 - 1.

(d) ijc - l)x2 + (A; + m + l)x + 3 = 0.

(e) x2 - X - 2(fc + m)a; = (fc + m) (x2 - 1).

(f) (fc + m)x2 + 2 (fc + m) + 1 = x2 - 2 x.

(g) (fc + m + l)x2 + (2 A: - m - l)x + 1 = 0.
(h) (2A; + m + 2)x2 + (4A: + 2m + 3)x + 3 = 0.

115. Sum and difference of roots. Let x^ and x^ be the roots of

x'' + bx + G = 0. (1)

Then (§ 95) x — x^ and x — x^ are factors, and their product
^'^ — (^1 + ^2)^ + ^1^2 is exactly the left-hand member of (1).
Consequently the equation

x^ -\- bx -\- c = x^ — (xi -{- Xz) X + X1X2
is true for all values of x. Hence by § 96

- (^1 4- X,) = b, (2)

X1X2 = c. (3)

We may state these facts in the

Theorem. The coefficient of x in the equation s(^-^hx-\-c=0*
is equal to the sum of its roots with their signs changed.
The constant term is equal to the product of the roots.

EXERCISES

1. Prove the statement just made from the expression for the roots in
terms of the coefficients (§ 89).

2. Form the equations whose roots are the following :

(a) 6, 1. (b) i, I (c) I, 3. (d) - i - 6.

(e) h h (f) -h+h (g) 2 + V3, 2 - V3. (h) - V3, V3.

* We should for the present exclude the case where 6^ - 4c<0, since the roots ar, and
a?j are then imaginary and we have not as yet dettned what we mean by the sum or
the product of imaginary numbers. We shall see later that the theorem is also true in
this case.

GRAPHICAL REPRESENTATION 107

3. If 4 is one root of ic^ — 3 x + c = 0, what value must c have ?
Solution : Let Xi be the remaining root.

Then by (1) _ (xj + 4) = - 3,

or xi = — 1.

By (2) c = Xi.4=:(-l)4 = -4.

4. Find the value of the literal coefficients in the following equations.

(a) x^-\-bx-9 = 0. One root is 3.

(b) 0:2 + 4 X + c = 0. One root is 2.

(c) ax2 + 3x-4 = 0. One root is 2.

(d) ax2 + 3x + 4 = 0. One root is 7.

(e) ax2 + 2 X + 6 = 0. One root is 6.

(f ) x2 + 6x + 4 = 0. One root is - 1.

(g) x2 - 6x - 6 = 0. One root is - 3.
(h) x2 + 6x + 6 = 0. One root is - 6.

(i) 2 x2 — 6 X — c = 0. One root is — 4.

(j) x2 — 6 X + c = 0. One root is double the other.

(k) x2 + c = 0. The difference between the roots is 8.

(1) x2 — 5 X + c = 0. One root exceeds the other by 3.

(m) x2 — 7 X + c = 0. The difference between the roots is 6.

(n) x2 — C X + c = 0. The difference between the roots is 4.

(o) x2 — 3 X + c = 0. The difference between the roots is 2.

(p) x2 — 2 X + c = 0. The difference between the roots is 8.

116. Variation in sign of a quadratic. It is often necessary to know the

sign of the expression

ax2 + 6x + c

for certain real values of x, and to determine the limits between which x may
vary while the expression preserves the same sign. We assume as usual that
a is positive.

Theorem L* If the discriminant of ax^ + hx + cis positive, the quadratic
is negative for all values of x between the values of the roots of the equation. For
other values of x {excepting the roots) the quadratic is positive.

* If a were negative, Theorem I would read as follows : If the discriminant is posi-
tive, the quadratic is positive for all values of x between the values of the roots of the
equation. For other values of x {excepting the roots) the quadratic is negative.

When a is negative Theorems II and III may be modified in an analogous manner.

108

\

In § 98 we found that when the discriminant of a quadratic equation is
positive the equation has two real roots. If two roots are real, the loop of
the graph of the equation ax^ + bx + cz=y cuts the X axis in two points

(§ 110) as in the figure. The roots are
represented by A and B, and any real
value of X between the roots is repre-
sented by a point P in the line J.JB.
Since the curve is below the X axis at
any such point, the value of y, i.e. of
the expression ax^ + &x + c for values
of X between the roots, is negative.

The value of the expression for any
value of X greater or less than both
roots is seen to be positive, since for
such points, for example Q and R, the graph is above the X axis.

Theorem II. If the discriminant of ax^ -\- hz + c is negative, the expression
positive for all real values of x.

When the discriminant is negative the entire graph of ax^ + &x + c = y is
above the X axis (§ 111), and consequently for any real value of x the corre-
sponding value of y, i.e. the value of ax^ + &x + c, is positive.

Theorem III. If the discriminant of ax^ + bx + c is zero, the value of the
expression is positive for all values of x except the roots of the equation
ax^ -\- bx + c = 0.

Hint. See example 16, p. 102.

We may restate these three theorems and prove them algebraically as
follows :

Theorem IV. If the discriminant of the quadratic ax^ + bx-\- cis positive,
the values of the quadratic and a differ in sign for all values ofx lying between
the roots, and agree for other values.

If the discriminant is zero or negative, the value of the quadraiic always
agrees with a in sign.

Case I. Since the discriminant is positive, the equation ax^+bx+c=(i
has two unequal real roots, as Xi and x^, of which we will assume Xi is the
greater, and we may write the quadratic in the form

ax2 + 6x + c = a (x — Xi) (x — X2).

Now for any value of x between Xi and X2 the factor x — Xi is negative,
while X — Xa is positive, which shows that the quadratic is opposite in sign
to a for such values of x. For other values of x both these factors are either
positive or negative, and for such values the quadratic is of the same sign
as a.

GKArillCAL KEPRESENTATION 109

Case II. Since the discriminant 6^ _ 4 ^c is negative and the roots are

* *!, * & ± V&2 - 4 ac .^ ^^ , -: .

of the form — — — — - , we may write the quadratic

2a

„ , , , / & V62 - 4 ac V , 6 . V62 - 4 ac\

V 2a 2a /\ 2a 2a /

Now for any value of x the expression lx-\ ) is positive, and since

- \ 2 a/
62 — 4ac is negative, 4ac — 6^ is positive; and we observe that tlie last
member of the equation has the same sign as a.

Case III. Since the discriminant is zero, the roots are equal and the
expression has the form

ax2 4- &x + c = a (x — Xi)2,

which has evidently the same sign as a, for any value of x.

EXERCISES

1. Between what values of x is the expression Vx^ — 5 x + 4 imaginary ?

Solution : The roots of x^ — 5 x + 4 = are 4 and 1.

The discriminant A = 62 _ 4 ^c = 25 — 16 = 9 is positive.
Thus by Theorem I or IV, if 1 < x < 4 f the expression under the radical
sign is negative and the whole expression is imaginary.

2. For what values of k are the roots of

(A:+ 3)x2 + A:x + l = (1)

(a) real and unequal ? (b) imaginary ?

Solution : a = k + S, b = k, c = 1.

A = b^-4ac = k^- 4{k + S) = k'^-ik- 12.

(a) If A > 0, the roots of (1) are real and unequal.
The roots of A;^ - 4 A: - 12 are A: = - 2 and 6
Then, by Theorem II, if

A; < - 2 or fc > 6, A > 0.

(b) By Theorem I, if - 2 < A; < 6, A < 0,
and the roots of (1) are imaginary.

* See 5 162. t Road " 1 is le6S than x wliich is lew than 4 " or " a; is between 1 and 4."

110 QUADRATICS AND BEYOND

3. Determine for what values of x the following expressions are negative,
(a) aj2 + 2x - 1/ (b) x^-Sx-h 4.

(c) x2 - 11 oj + 10. (d) x^-15x-{- 60.

(e)-x2-2x + l. {i) - x^ -\- 7 X + SO.

4. Determine for what values of k the roots of the following equations
are (a) real and unequal, (b) imaginary.

(a) 3 A;x2 - 4 X - 2 = 0. (b) aj2 + 4 fcx + A:2 + 1 = 0.

(c) x^-{-9kx + 6k + l = 0. (d) x2 + {3k + l)x + 1 = 0.

(e) (A;2 + 3)x2 + A;x - 4 = 0. (f) 2 x2 - 4x - 2 A; + 3 = 0.

(g) fcx2 + (4A; + l)x + 4 A: - 3 = 0. (h) {k - l)x2 + 6kx -{- 6k + i = 0.
(i) {k - l)x2 4- {2k+l)x + A: + 3 = 0.

CHAPTEE X

SIMULTANEOUS QUADRATIC EQUATIONS IN TWO
VARIABLES

117. Solution of simultaneous quadratics. A single equation
in two variables, as x^ -\- y^ = 5, is satisfied by many pairs of
values, as (1, 2), (Vl> V|)> (2, 1), and so on, though there are
at the same time numberless pairs of values that do not satisfy
it, as (0, 1), (1, 1), (2, 3). Thus the condition that (x, y) satisfy a
single quadratic equation imposes a considerable restriction on
the values that x and y may assume. If we further restrict the
value of the pair of numbers (cc, y) so that they also satisfy a
second equation, the number of solutions is still further limited.
The problem of solving two simultaneous equations consists in
finding the pairs of numbers that satisfy them both.

118. Solution by substitution. In this method of solution the
restriction imposed on (x, y) by one equation is imposed on the
variables in the other equation by substitution.

Example. Solve 2x^-\-y'^-l, (1)

x-y = \. (2)

Solution : Equation (2) states that x = 1 + y. Thus our desired solution is
such a pair of numbers that (1) is satisfied and at the same time x is equal
to y + 1.

If we substitute in (1) 1 + y for x, we are imposing on its solution the
restriction implied by (2).

Thus 2(l + 2/)2 + 2/2=,i^

or \ 3 2/2 4- 4 y + 1 = 0.

The roots are y = — 1, y = — i.

Corresponding to ?/ = — 1 we get from (2) x = 0.
Corresponding to y =— \ we get from (2) x = |.
Thus the solutions are (0, — 1) and (|, — \).

Ill

112 QUADRATICS AND BEYOND

EXERCISES

Solve the following :

xy = 4.

2.

x-y = 6,
xy = 36.

■ x2 + 2/2 = bxy.

xy — x = 0.

5.

x-hy = a,

x2 + 2/2 =, 5.

x2 4-2/2 = 50,
• 9x + 7 2/ = 70.

'• x-Sy = 0.

8.

2x-3y = 4,
x2 - y2 = 0.

9 xy = 12,

2x + 3?/ = 18.

10 x:y = 9:4,
^"- a;:12 = 12:y.

11.

X2:y2=:a2:62^
a — X = b — y.

12 5x2 + 2/ = 3xy
2x-2/ = 0.

*"• 2x-32/ = 0.

14 (a^ + y)C

z-2y) = T,
3.

-g 3x2-4y = 6x-
3x+42/ = 10.

2y%

16 x2 + ,=
x:y = 2

2/2 + X - 18,
:3.

^^ ax-by = cy, ^g x2 + 2 X2/ + 2/2 = 7 (x - y),

a2x2 — 62y2 _ acx2/ + to2. ' 2 X — 2/ = 5.

^g ax2 + (a-6)x2/-&2/=^=c2, ^q 2x2-5x2/+2/2+10x+12y=100,

■ (x + 2/) : (X - 2/) = a : 6.

2j 7(x + 5)2 -9(2/ + 4)2 = 118,
* X - 2/ = 1.

x2 + 2/2 = 130,
23. X + y ^

x-y

2X-2/ + 1 8

25.

x-22/ + l~3'

x2 - 3 X2/ + 2/2 = 5.

27.

X2/ + 72 = 6(2x + 2/),
X 2

V 8

4X+2/-1 4X+2/-12

29.

2X+2/-1 2X+2/-12

3x + y = 13.

31.

10 9'
x + 2 + 2/-l = ''
2 4

= 2f,

2x-32/ = l.

22.

x2 + 2/2 = a2,
X m

2/ n

24.

x2 + 2/ + 1 3
2/2 + X + 1 2

x — y = l.

26.

1 + X + X2^3

1 + 2/ + 2/^
a; + 2/ = 6. •

28.

2/ + 6 X

X — 2/ = 4.

30.

«(x-2/)-52/ = 6
x-2/

32.

? + ? = 3,

x 2/

^^Tri = ;- 5(2/-l) = 2(x + l).

SIMULTANEOUS QUADRATIC EQUATIONS 113

119. Number of solutions. We have proved (p. 42) that two
linear equations have in general one and only one solution.

Theorem. A quadratic equation and a linear equation ham
in general two and only two sohUions.

If the linear equation is solved for one variable, say x, and this
is substituted in the quadratic equation, we get a quadratic equa-
tion to determine all possible values of the other variable (i.e. y),
which must in general be two in number (§ 98). To each one of
these values of y will correspond one and only one value of x,
thus affording two solutions of the pair of equations.

EXERCISES

1. When may, as a special case, a quadratic and a linear equation
have only one solution?

2. When may a quadratic and a linear equation have imaginary
solutions ?

3. rind the real values of k for which the following equations have
(1) only one solution, (2) imaginary solutions.

x2 + y2 = 16, (1)

^^^ x-y = k, (2)

Solution : x = y + k.

Substitute in (1), {y + fc)2 -{- y"^ = 16.
or 2 y2 4- 2 A:y + A;2 - 16 = 0.

As in § 98, a = 2; b = 2k; c = k^ -16.

Hence A = b'^ - Aac = ^k^ - Sk^ + 12S = - Ak^ + 128. •

(1) A = when 4k^ = 128,

or k = ± 4 -^2. There is then only one solution.

(2) A < when k^ > 32. The solution is then imaginary.

^^ 2x-y = k.

x^-y^ = 9,
^^^ x-2y = k.

2x2 + 3y2 = 6,
^J^ x-ky = \.

x + ky = b,

^""^ (C2 + 2/2 = 5.

^ y-x = k,

^ ^ x^y = k.

x2 + 2/2 = 25,
(') 4x-3y = k.

114 QUADRATICS AND BEYOND

1

. 120. Solution when neither equation is linear. In the exam-
ples previously given one equation has been linear and the other
quadratic in one or both variables. Often when neither of the
original equations is linear a pair of equivalent (p. 41) equations
one or both of which are linear may be found. These latter equa-
tions may be solved by substitution.

EXERCISES

Solve the following equations.

When neither equation is linear, we can often obtain by addition an
equation from which by the extraction of the square root a linear equation
may be found.

1. x2 + 2/2 = 17, (1)

xy = 4. (2)

Solution: x^ + y^ = 17 (3)

Multiply (2) by 2, 2xy = 8 (4)

Add x2 + 2 icy + 2/2 ^ 25

Extract the square root, a; + y = ± 5. (5)

Subtract (4) from (3), x2 - 2 xy + 2/2 = 9.

Extract the square root, x — y =±S. (6)

Solve (5) and (6) as simultaneous equations,

X + y = ±6,

x~y = ±S.

x = +i, +1, -1, -4.
2/ = +l, +4, -4, -1.
Thus the solutions are four in number, (4, 1), (1, 4), (—1, — 4), (— 4, —1).

The following exercise affords another case where a linear equation may
be found by addition and extraction of the square root.

2. x^-hxy = 6, (1)
xy + y^ = 10. (2)

Solution : Add (1) and (2),

x^-^2xy + y^ = 16.
Extract the square root, « + 2/ = ± 4.

Substitute in (1), x2 + x ( ± 4 - x) = 6,
x2±4x-x2=:6,

X = ± f = ± 1^.

Substitute in (4), 2/ = 2^, — 2|.

Thus our solutions are (- |, - 2|), (f, + 2J).

SIMULTANEOUS QUADRATIC EQUATIONS 115

When neither of the original equations is quadratic, we can often find by
division an equivalent pair of equations one of which is linear and the other
quadratic, as in the following exercise.

3. x3 + 2/8=12, (1)

X + 2/ = 2. (2)

Solution: Divide (1) by (2),

x2 - ajy + y2 ==: 6. (3)

Square (2), x^ + 2 xy + y^ = 4

Subtract, — 3 xy = 2

xy=-l (4)

Solve (4) with (2) by substitution.

When the sum of the exponents of the variables is the same in every
term, the equation is called homogeneous.

Thus, z'^ + xy^O, 2x^y -3zy^ - 'iz^ - 3y^ = 0.

When one equation is homogeneous and the other either linear or quadratic
we may solve them as follows :

4. 6x2-7xy + 2y2 = o, (1)

x2 - y = 4. (2)

Solution : Divide (1) by y^,

Let- = 2,« 622_7z + 2 = 0.

y

Solve for z, z = | or |.

Thus ?=:ior? = ?.

y 2 y 3

Solve (2) with 2 x = y and 3 x = 2 y.

When both equations are homogeneous except for a constant term we may
solve as follows :

5. x2-x2/ + 2y2 = 4, (1)
2x2-3x!/-2?/2 = 6. (2)

Solution : Eliminate the constant term by multiplying (1) by 3 and (2) by 2,
3x2-3x2/+ 6?/2 = i2, (3)

4 x2 - 6 xy - 4 y2 = 12 (4)

Subtract (3) from (4), x2 - 3 xy - 10 y2 = o

* "We observe that y ^0. For if y = were a value that satisfies equation (1), x =
would correspond. But (0, 0) does not satisfy (2); thus y = Ois not a value that can occur
in the solutions of the equations.

116 QUADRATICS AND BEYOND

X

Divide by y^ and let - = 2, where y ^d,

y

22 _ 32 _ 10 = 0. (S)'

Factor, (z - 6) (2; + 2) = 0.

The roots are _ = 6, - = - 2. (6)

y y
Solve (6) with (1).

"When one equation is quadratic in a binomial expression we may solve as
follows :

6. x-y

-Vx-2/ = 2, (1)

x^-ys = 2044. (2)

Solution: Let

Vx - ?/ = z.

Then (1) becomes

z2 - z = 2.

Solving for z.

z = 2 or - 1.

Thus
or

Solve (3) with (2) as in exercise

x-2/ = n (3)
x-y = lj ^ '

3.

7. x^ + y^ = xy = X -\- y.

8. X3 + 2/3 =7x2/ = 28(x + 2/).

2/3 + x^y = 4.

x2^ = a,
^"- X2/2 = b.

- x{y-l) = 10,
' y{x-l) = 12.

12 a^2 + 2/2-a,
' xy = b.

^^ x^y + xy^ = a,

x^y — xy^ - b.

.. x + X2/=:35,
• 2/ + X2/ = 32.

-g x{x^-{-y^) = 7,
y{x^-\-y^) = l.

2x2-32/2 = 6,
• 3x2-2 2/2 = 19.

3x2-22/2 = 16.

-J. 5x2 + 2^2 = 22,
3x2-52/2 = 7.

IQ ic2 + ccy + 2/2 = 2,
• x2 - X2/ + 2/2 = 6.

20 x + X2/ + 2/ = 5,

■ a;2 + X2/ + 2/2 = 7.

Hint. Eliminate x2 or 1/2 as if _ . . o «

the equations were linear equa- 21. "^ » ^ i'j

tions in x2 and y^. X2/ = 2 x - 2/ + 9.

22 «2_x2/ + 2/2 = 37, (x + 2/)(8-x) = 10,

• x2 - j/2 = 40. ■ (X + 2/) (5 - 2/) = 20.

24 (»2 + 2/2)(x + 2/)=6, 25 (^ + 2/)^ = 3x2 - 2,

' «y(x + 2/) = - 2. * • (X - 2/)2 = 32/2 - 11.

117

26.

X + y/x^y = a,
y + ^x?/2 = 6.

28.

X + y = 58,

Vx + Vy = 10.

30.

X + 2/ = 3.

32.

4x*-9y2 = o,

^ —O ■ ..O O / 1 -A

34.

38.

40.

42.

4x2 + 2/2 = 8(x + y).
3x2/-2(x + y) = 28,
2x2/-3(x + 2/) = 2.

x2 + ?/2 + X + y = 18,
x2 - 2/2 + X - y = 6.

3x2- 2 2/2 = 6(x-2/),
X2/ = 0.

x2-X2/ + 2/2 = 13(x-2/),
xy = 12.

Vx(l-y) +V2/(l-ic) = a,

2/

^^ Vl-x2Vl-2/2 + X2/ = |l,
• x-y = ll.

X _y _ 16
46. y X " 16'

3x2 + 6y2 = i20.

X V^ + 2/ Vy _ 1
48. xy/x — yy/y 2

x3 - 8 = 8 - y3.

V2/ — Va — X = Vy -X,
50. Vy — X + Va — x _ 5

52.

Va — X

1 , 1_3

i + i = l
x2 y« 4

27.

29.
31.

33.
35.
37.
39.
41.

43.
45.

47.
49.

51.
53.

xVx + y = 3,
y Vx + y = 1.

■y/x +y/y = a,
X + y = 6.

Vx — Vy = 2,
(x + y) Vxy = 610.

Vx-6 + Vy + 2 = 5,
X + y = 16.
xy + xy-i = x2 + y2,
xy-xy-i = 2(x2 + y2).

x-2 + 2 y-2 = 12,

x-2 - x-iy-i + y-2 = 4.

5c2 + y2_5(a; + y)^8,
x2 + y2-3(x + y) = 28.

2x2-3xy + 6y- 6 = 0,
(x-2)(y-l) = 0.

V6-3x + x2 + V6-3y + y2 = 6,
X + y = 3.

X y '

y = .3.

a y _ 25

y '^x~12*
x2 - y2 = 28.

V 5x /x +

a; + y \ 6a
icy - (x + y) = 1.

^ + ?^ = 2,
a2 62

bx + ay _m

bx — ay n

x8 _ y3 _ 16
y X 2 '

- -1=-.
y x ~ 2'

y ^ 3 V2
X 2 '

Hint. Let - = u, - = v.

a ' y

118 QUADRATICS AND BEYOND

X -1 a—1

2,
54. ) ^i 55.

3.

y-\ b-
x3 - 1 a3

yS _l 53 _ 1

lift ^ + ^ 57 ^^^

^ 2 x + l 221

1 1_5 1_?-1

5g x'^y~6' 59 "^ ^~ '

X + 22/ + I2 x +

xy = 2.

PROBLEMS

1. Two numbers are in the ratio 5 : 3. Their product is 735. What are
the numbers ?

2. Divide the number 100 into two parts such that the sum of their

squares is 5882.

3. The sum of the squares of two numbers increased by the first is
205 ; if increased by the second the result is 200. What are the numbere ?

4. The diagonal of a rectangle is 85 feet long. If each side were longer
by 2 feet, the area would be increased by 230 square feet. Find the length
of the sides.

5. The diagonal of a rectangle is 89 feet long. If each side of the rec-
tangle were 3 feet shorter the diagonal would be 85 feet long. How long
are the sides ?

6. The sum of two numbers is 30. If one decreases the first by 3 and
the second by 2 the sum of the reciprocals of the diminished numbers is ^.
What are the numbers ?

7. The sum of the squares of two numbers is 370. If the first were
increased by 1 and the second by 3, the sum of the squares would be 500.
What are the numbers?

8. A number of persons stop at an inn, and the bill for the entire party
is \$24. If there had been 3 more in the party, the bill would have been
\$33. How many were in the party and how much did each pay ?

SIMULTANEOUS QUADRATIC EQUATIONS 119

9. A fruit seller gets \$2 for his stock of oranges. If his stock had con-
tained 20 more and he had charged f of a cent more for each, he would have
received \$3 for his stock. How many oranges had he and how much did he
get apiece for them ?

10. A man has a rectangular plot of ground whose area is 1250 square
feet. Its length is twice its breadth. He wishes to divide the plot into a
rectangular flower bed, surrounded by a path of uniform breadth, so that
the bed and the path may have equal areas. Find the width of the path.

11. In going 7500 yards one of the front wheels of a carriage makes 1000
more revolutions than one of the rear wheels. If the wheels were each a yard
greater in circumference, the front wheel would make 625 more revolutions
than the rear wheel. What is the circumference of the wheels ?

12. A man has |539 to spend for sheep. He wishes to keep 14 of the
flock that he buys, but to sell the remainder at a gain of \$2 per head.
This he does and gains \$28. How many sheep did he buy and at what
price each?

13. A man buys two kinds of cloth, brown and black. The brown costs
25 cents a yard less than the black, and he gets 2 yards less of it. He
spends \$28 for the black cloth and \$25 for the brown. How much was each
a yard and how many yards of each did he get ?

14. A man left an estate of \$54,000 to be divided among 8 persons, namely,
his sons and his nephews. His children together receive twice as much as
his nephews, and each one of his children receives \$8400 more than each one
of his nephews. How many sons and how many nephews were there ?

15. A and B buy cloth. B gives \$9 more for 60 yards than A does for
45 yards ; also B gets one yard more for \$9 than A ddes. How much does
each pay?

16. A sum of money and its interest amount to \$22,781 at the end of a
year. If the sum had been greater by \$200 and the interest :^ of 1 per
cent higher, the amount at the end of the year would have been \$23,045.
What was the sum of money and what was the interest ?

17. If one divides a number with two digits by the product of its digits,
the result is 3. Invert the order of the digits and the resulting number is in
the ratio 7 : 4 to the original number. What is the number ?

18. What number of two digits is 4 less than the sum of the squares of
its digits and 5 greater than twice their product ?

19. Increase the numerator of a fraction by 6 and diminish the denomi-
nator by 2, and the new fraction is twice as great as the original fraction.
Increase the numerator by 3 and decrease the denominator by the same, and
the fraction goes into its reciprocal. What is the fraction ?

120 QUADRATICS AND BEYOND

121. Equivalence of pairs of equations. In the theorems of
this section the capital letters represent polynomials in x and y,
and the small letters represent numbers not equal to zero.

I

Theorem I. The pairs of equations
A

are equivalent.

If (ari, 2/1) be a pair of values that satisfy (1), then when x and
y in 5^ are replaced by Xi and yi the equation B"^ = i^ is a numer-
ical identity. These values (xy, y^ must then satisfy one of the
equations -S = ± ^, for if they did not, but only satisfied the equa-
tion say B = c when c =^ ±b, then the hypothesis that B^ = }p- is
satisfied by (cci, ?/i) would be contradicted.

Conversely^ any pair of values that satisfy B = -^h evidently
satisfy B"- = b\

This theorem is used, for instance, in exercise 2, p. 114, and justifies the
assumption that

are equivalent pairs of equations.

Theorem II. The pairs of equations

are equivalent.

li A = a and B = h are satisfied by a pair of numbers (xj, y^),
we multiply the identities and obtain AB = ab.

Conversely f if A = a, AB = ab are identically satisfied by a
pair (xi, yi), since a ^ we can divide the second identity by
the first and obtain B = b. Thus if (cci, yi) satisfy one pair of
equations they satisfy the other pair.

This theorem is assumed in exercise 3, p. 115, to show that

a;« + w« = 12 ^ , a;2 - XM + 7/2 = 6 1 . , ^

, ^ y and . " ~r ^ ^ are equivalent.

SIMULTANEOUS QUADRATIC EQUATIONS 121

Theorem III. The pairs of equations

^:^}a) ^f!^!:^K^)

B = 0]^ ' cA-\-dB =
are equivalent where a, b, c, and d are numbers such that

ad — be ^ 0.

If (xi, yi) satisfy (1), evidently they also satisfy (2). Thus all
solutions of (1) are among those of (2).
Conversely J if (xi, ^/i) satisfy (2), then

__bB__dB
a c

Thus (ad — hc)B = 0.

Thus since (ad — bo) =^ 0, .

^ = 0.

Similarly, ^ = 0.

This theorem has been assumed in exercises 1, 2, 3, 6, p. 114. In 1, for example,
it is necessary to show that

are equivalent. In this case a=c=l, b = — d = 2. Thus ad — be = — 4: ^t 0.

122. Incompatible equations. When a pair of simultaneous
equations can be proven equivalent to a pair of equations which
contradict each other or are absurd, they are incompatible and
have no finite solution.

(1)
(2)

Example 1.

xy = 1

Subtract,

xy = -1
0= 2

Example 2.

x'^ + y^ = 4,
4x2 + 42/2 = 49.

Multiply (1) by 4,
Subtract,

4x2 + 42/2 = 16

4x2 + 42/2 = 49

= 33

122

123. Graphical representation of simultaneous quadratic equa-
tions. Every equation that we have considered may be rep-
resented graphically by plotting
in accordance with the method
already given (p. 93).

The solution of simultaneous
equations is represented by the
points of intersection of the cor-
responding graphs.

Thus the equations

x'^,f = 25,
2xy = 9
have the solutions

a;=±2±

V34

y==F2±

V34

or

X = 4.9, .9, - .9, - 4.9,

y

.9, 4.9, - 4.9, - .9.

These equations have as their graph the preceding jfigure.
The equations

ic2 + 2£cH-4?/ + l = 0,
x-f22/ + 4 = 0,

which have the solutions

a; = ± Vt = ± 2.6,

y = - 2 =F V^ = - 3.3 or -

Y'^

- ^ -

^^

^N, X

Z '^

^J\ -

/

^\r -

r

%>

7

L^

!-—-

5:

r>

~fcM —

^^ ^

, Ny

1 /T

"\ V

I ^

^ I^

-\ ^ °

-4 t*

-^^^-

J t

_ ^^

7

"^

have as their graph the figure shown
above.

As in the case of linear equations,
incompatible equations afford graphs
which do not intersect. Thus the graph
of the equations in example 2, p. 121, is
found to be two concentric circles, as
is shown in the adjacent figure.

123

Simultaneous equations which have
imaginary solutions also lead to non-inter-
secting graphs (p. 101).

Thus the equations

a;2 + 2/' = 4,
have the adjacent figure as their graph.

. .^.-

.

^,

V :

■ /"=VlXI 1 1

1 >L f

J hs'

^^^/| 1 1 U"

EXERCISES

1. Inteipret the graphical meaning of equivalent pairs of equations.

2. Plot and solve x^ + ?/2 = 2,

X + 2/ = 2.
What general statement concerning the graphical meaning of a single
solution of quadratic and linear equations does this example suggest ?

3. Plot and solve the following :

(a)

25.

X2 + 2/2

4x2 + 9^2 = 144.

(b)

(c)

x2 + 2/2 = 25,

x2 + y2 = 25,

4x2 -8x + 92/2 = 140.

5x2 + 2/2 = 25.

What general statement concerning the graphical interpretation of four,
three, or two real solutions of equations do these examples suggest ?

4. State the algebraical condition under which two quadratic equations
have four, three, two, or one real solutions (see p. 113).

5. Plot and solve the following :
x2 + 2/ = 0, „ . a;2 + 2/2 = 9.

32/ = 0.

(a)

(c)

(e)

X2

4x-22/ = 3,

^.2 I 2/2
y^' x2 - y2 = 0.

xy

0.

^ ' X2 + y2

16.

4x2 + 92/2 = 36,

x2 = — 4 ?/.

(f)

xy = 1,
2x-Sy = lS.

124. Graphical meaning of homogeneous equations. Consider for example
the homogeneous equation

3x2 -10x2/ -82/2 = 0. (1)

If we let z = - , we get

X

or
or

3 _ 10 z - 8 z2 = 0,

8^2 + 102-3 = 0,

(4z-l)(2z + 3) =0.

124

The roots are

2 = 1 and 2 = - |.

Thus

y=i and y=-l,

a; 4 X 2

4y-x = and 3x + 22/ = 0.

or

These equations represent two straight lines through the origin which
taken together form the graph of equation (1). This
example may obviously be generalized : Any homo-
geneous equation of the form ox^ + hxy + cy"^ = with
positive discriminant represents two straight lines
through the origin. Such an equation is equivalent
to two linear equations.

In an example like 5, p. 115, we obtain in place
of the given pair of equations a pair of equivalent
equations one of which is homogeneous and the other
of which is factorable. We can learn the graphical
meaning of tl;is method of solution by studying a
particular case. Consider for example the equations :

jc2 + 2x2/ + 7?/2 = 24, (1)

2x^-xy-y^ = S. (2)

By eliminating the constant terms we obtain the product of the two
equations x + y = and x — 2 y = 0. Thus the problem of solving (1) and
(2) is replaced by that of solving the two following pairs of equations :

x2 + 2X2/ + 72/2 = 24, (^) (2) . (2)

x + y = 0,
or x2 + 2x2/ + 73/2 = 24,

X - 2 2/ = 0.

The graphical meaning of this
method of solving the equations (1)
and (2) is seen in the fact that the
problem of finding the points of inter-
section of the graph of equation (1)
with that of (2) is changed to that of
finding the intersection of the graph of (1) with a pair of straight lines.
This appears in the figure where the curves and lines are numbered as
above. The closed curve represents (1).

CHAPTEE XI
MATHEMATICAL INDUCTION

125. General statement. Many theorems are capable of direct
and simple proof in special cases, while for the general case a
direct proof is difficult and complicated.

If we ask whether ic" — 1 is divisible by x — 1, it is easy to
make the actual division for any particular value of n, as n = 2
or n = 3. But if x^ — 1 is shown divisible by a; — 1, we are no
wiser than before concerning the divisibility of x^ — 1. Suppose,
however, we can prove that the divisibility for ti = m -f 1 follows
from that for n = m, whatever value m may have. Then since
we have established the fact by direct division for n = 3, we may
be assured of the divisibility for ?i = 4, then for n = 5, and so on.

Now x'^+^ — l=x(x"' — l)-\-(x — l)

is identically true. If ic — 1 is a factor of cc"* — 1 for a given value
of m, it is a factor of the right-hand member and consequently a
factor of the left-hand member (§ 69), which was to be proved.
Thus the divisibility of x"" — 1 hjx — 1 is established for any
integral value of n greater than the one for which the division
has actually been carried out.

To complete the proof of a theorem by mathematical induction
we must make two distinct steps.

^-^^^st Establish the theorem for some particular case or cases,
preferaUy for n =1 and n = 2.

Second. Show that the theorem for n = m -{- 1 follows from
its assumed validity for n = m.

Example. Prove that the sum of the cubes of the integers

from 1 to 71 is SK^(^ + 1)]S'-

To prove that 1« -f- 2» 4- 3* + • • • -F 7i» = J^[7i(n + 1)]^.

125

126 QUADRATICS AND BEYOND

First. This theorem is true for n = 1.

For 1» = 1 = J ^ [1 (2)] p = 12 = 1

The theorem is also true for n = 2.

For l» + 2» = 9= J^[2(2 + l)]P = G-6)2=32 = 9.

Second. Assume the theorem for n = m,*

1« + 2« + . . . + w« = J J[m(m + 1)]P.

Add (m + 1)^ to both sides of the equation,

1« + 2« + . . • + m« + (m + 1)« = ^[m(m + 1)] p + (m + 1)*

= [(^m)2-|-m + l](m + l)2

= ( 4 ) (^ + ly

= [i(mH-l)(m + 2)]S
which is the form desired, i.e. m + 1 replaces m in the formula.

EXERCISES
Prove by mathematical induction that

1. 1+3+6 + .. .+(2n-l)=n2.

2. 2 + 22 + 28 + . . . + 2" = 2(2» - 1).

3. 3 + 6 + 9 + ... +3n = f n(n + l).

4. 12 + 22 + 32 + ... + n2 = ^n(n + l)(2n + l).

5. 13 + 28 + 33 + . . . + n3 = (1 + 2 + 3 + . . . + n)2

6. 42 + 72 + 102 + . . . + (3 n + 1)2 = ^ n(6 n2 + 16n + 11).

7. X" — y" is divisible by x — y for any integral values of n.

8. x2» — 2/2" is divisible by x + y for any integral values of n.

9. 1.2 + 2-3 + 3. 4 + 4.6 + . .. + n.(n + l) = ^n(n + l) (n + 2).

10. 1 • 1 + 2 • 32 + 3 . 62 + ... + n(2 n - 1)2 = ^n(n + l)(6n2 - 2n - 1).

11. 1.2.3+2.3.4+3.4.6 + ...+n(n+l)(n+2) = in(?i+l)(n+2)(n+3).

12. (1« + 28 + 38 + • . . + n8) + 8(16 + 26 + 35 + . . . + n6)

= 4(1 + 2 + 3+ ... +n)8.

* This statement does not imply that we assume the validity of the formxila for any
values for which it has not yet been established, but only for values of m not greater
than 2.

MATHEMATICAL INDUCTION 127

14.

Ill 1 _ ^

1.2'2.3'3.4' 'n-(w + l) n + 1

15.

o

16.

2.6 + 3.6 + 4.7 + ... + (« + l)(. + 4) = "<» + ''g><'' + «V

17. 2.4 + 4.6 + 6.8 + -.. + 2n(2n + 2) = ^(2n + 2)(2n + 4)

o

18. A pyramid of shot stands on a triangular base having m shot on a
side. How many shot are in the pile ?

CHAPTEE XII

BINOMIAL THEOREM

126. Statement of the binomial theorem. When in previous
problems any power of a binomial has been required we have
obtained the result by direct multiplication. We can, however,
deduce a general law known as the binomial theorem, which gives
the form of development of (a + hy, where n is any positive integer
and a and h are any algebraical or arithmetical expressions. This
law is as follows :

(a 4- ^»)« = a« + ^ a«- ^b + ^ ' ^f ~ ^^ a^-^'b^ + "- + b\
From this expression we deduce the following
EULE FOR THE DEVELOPMENT OF (a + bf.

The first term is a".
n
1

To obtain any term from the preceding term, decrease the
exponent of a in the preceding term by 1 and increase the
exponent of b by 1 for the new exponents. Multiply the coefficient
of the preceding term by the exponent of a, and divide it by the
exponent of b increased by 1 for the new coefficient.

Remark. In practice it is usually more convenient first to write down all the
terms with their proper exponents, and then form the successive coefficients.

EXERCISES

Verify by multiplication the rule given for the following :

1. (a + 6)». 2. (x - 2/)8.

3. (2a + 36)*. 4. (v^+Vi^)'.

5 (2 a -6)4. 6. {x-y/y)\

7. (3a -2 6)8. 8. (a-ia; + 6-iy)4.

128

BINOMIAL THEOREM 129

127. Proof of the binomial theorem. We have already stated
that / -j \

{a + ^)« = a" + 7 a--^b + ""^ ~ ^ a^-'b"" + ••. (1)

and have seen that it is justified for every particular case that
we have tested. By complete induction we now prove this
theorem when ti is a positive integer.

First. Let n = 2.

That is, (a -{- by= a^ -{- 2 ab + b\

This expression evidently obeys the law as stated in (1).
Second. Assume the theorem for n = m.

That is, (a + &)- = a- 4- ^ a^-^h + '^^'^ -^) ^m-2^2 + .... (2)

Multiply both members hj a -\-b,

(a 4- Z>)"'+i = a'«+i ^ja'^h^j a'^-^b'^

z

Z
This expression is identical with (2) except that (m + 1)
replaces m. Hence the theorem is established so far as the
first three terms are concerned.

128. General term. Though we have stated the binomial
theorem for a>§eneral value of ri, we have only established the
exact form of the first three terms.

Let {a + by = a'* + c^a^'-^b + Cga"" ^'^ ^ . ,

We note that the sum of the exponents of a and 5 is w in any
term of the development of {a -\- by. Also the exponent of b in
the (r -\- l)st term is r.

We have already seen that

n n(n — l)

and that the first three terms are

J. ± • z

130 QUADKATICS AND BEYOND

respectively. This indicates that the coefficient of the next term

will be — ^^ — - — -^r-^ and in general that the coefficient of the

(r + l)st term has the form

_ n(n-l)---(n-r + l)

l-2...r ' W

which is in fact the form that our rule (§ 126) would afford for
any particular value of r.
This affords the following

Rule. The (r + l)st term of {a + hf is

n{n-l)-"{n-r + l) ^„_ , ^,
1.2'- r

The form of the coefficient may be easily remembered since the
denominator consists of the product of the integers from 1 to r,
while the numerator contains an equal number of factors consist,
ing of descending integers beginning with n.

For any particular values of n and r we could easily verify the
rule by direct multiplication. For the rigorous proof see p. 178.

EXERCISES

Develop by the binomial theorem :

/_a_ _ ^^Y

Solution

6-5-4-3 (_±y(_ VxV 6.5.4.3-2 / a V/ VaV

1.2.3.4VV^/V a2/ l-2.3.4.6VVi/\ aV
6. 5-4. 3.2.1 / V^y
1.2.3.4.5.6V a2/

x« x2 X a3 a6 a* "^ a^a"

2. (f-fa)6. 3. (Va + v'ft)'.

BINOMIAL THEOREM 131

10. (1 + V^y - (1 - V^)'. 11. ( Vx + v^)*+ ( Vx - Vy)*.

(2 X 3 v\^^
1 ) •
Sy 2x7

Solution : n = 10, r + 1 = S.

The (r + l)st term of (a + 6)" is (§ 128)

n(n-l)-'-(n-r + 1)

1-2. -r
In this case we get

10-9-8.76-5.4 (^xy/Syy

1.2.3.4-6-6.7 '\3y)\2x)

33y3 27x7 24X*

_ 120-81 y^_ 1215y4
16 'x*~ 2x* '

(1\13
a + -) .

14. Find the 6th term of (— - ^V^.

\2y xj

15. Find the 8th term of ( — - ^ ) .

\y X /

16. Find the 6th term of l2aVb ) .

V 2aVb/

17. Find the 7th term of (^ - ^) •

18. Find correct to three decimal places (.9)^
Solution; (.9)8 = (1 - .1)8

8^7^6^ ^

1.2.3.4^ / V ' -r
= 1 - 8 • 0.1 + 28 • 0.01 - 66 . 0.001 + 70 • 0.0001
= 1 + 0.28 + 0.0070 - 0.8 - 0.056
= 1.2870 - 0.856 = .431.

In this exercise any terms beyond those taken would not affect the first
three places in the result.

132 QUADRATICS AND BEYOND

Compute the following correct to three places :

19. (1.1)10. 20. (2.9)8. 21. (.98)11. 22. (1.01)«^

23. (1^)8. 24. (1^)10. 25. (98)8. 26. (203)5.

27. In what term of (a + 6)2o does a term involving ai* occur?

28. For what kind of exponent may a and h enter the same term with
equal exponents ?

29. For what kind of exponent is the number of terms in the binomial
development even ?

30. Find the first three and the last three terms in the development of

(Z, . 1 \24

CHAPTEB XIII
ARITHMETICAL PROGRESSION

129. Definitions. A series of numbers such that each numbei
minus the preceding one always gives the same positive or
negative number is called an arithmetical series or arithmetical
progression (denoted by A.P.).

The constant difference between any term and the preceding
term of an A.P. is called the common difference.

The series 4, 7, 10, 13, • • • is an A.P. with the common difference 3. The series
8, 62, 5, 32, • • • is an A.P. with the common difference — §. The series 4, 6, 7, 9,
10, • • • is not an A.P.

EXERCISES

Determine whether the following series are in A.P. If so, find the common
difference.

1. 6, m If, .... 2. 27, 22^, 18, ....

3. 6, 4^ 3, H, .... 4. 6, -2, -8, ....

5. VI, V2, 3 VI, •••. 6. 8, 5|, 3|, If, ....

- 1 2 4 V2-I V2 1

'• V2 V2 V2 2 2 2(V2-1) *

9.3, -^, -3f, -6f, .... 10.^ ^^ + 2 -^^

6 6(V3-4)

130. The nth. term. The terms of an A.P. in which a is the
first term and d the common difference are as follows:

a, a -i- d, a -\- 2d, a -{- 3dj '". (1)

The multiple of d is seen to be 1 in the second term, 2 in the
third term, and in fact always one less than the number of the
term. If we call I the nth. term, we have

I = a -{■ (^n — 1) d.

133 ^ .., a

134 QUADRATICS AND BEYOND

We may also write the series in which I is the nth term as*

follows :

a, a-{- dy a + 2d, • • • , I — 2d, I — d, I.

131. The sum of the series. We may obtain a formula for
computing the sum of the first n terms of an A.P. by the following

Theorem. The sum s of the first n terms of the series
a, a -^dy • • •, I — d, I is

By definition,

s = a-\-(a + d) + (a + 2d)-\ -]-{l - 2d)-\-(l - d)+ I. (1)

Inverting the order of the terms of the right-hand member,

s =. I + {l - d) + {I - 2 d) -^ ' ■ - + {a + 2 d) + (a -\- d)+ a. (2)
Adding (1) and (2) term by term,
2s^{l-\-a)+{l + a)-\-{l + a) + .'-^{l-\-a) + (l-\-a) + {l + a)

z=n{a-\- 1).
Thus s = ^(a-\-l).

132. Arithmetical means. The terms of an A.P. between a
given term and a subsequent term are called arithmetical means
between those terms. By the arithmetical mean of two numbers
is meant the number which is the second term of an arithmetical
series of which they are the first and third terms. Thus the

arithmetical mean of two numbers a and h is — r— > since the

numbers a, — ^r— > h are in A.P. with the common difference — — •

The two formulas

« = a + (»i - 1) d, (I)

« = -(« + 1) (II)

contain the elements a, I, s, n, d. Evidently when any three are
known the remaining two may be found by solving the two equa-
tions (I) and (II).

AKITHMETICAL PROGRESSION 135

EXERCISES

1. Find the 16th term and the sum of the series 4, 2, 0, — 2, • • ..
Solution: n = 16, a = 4, d = 2 - 4 = - 2.

Z = a + (n - l)d = 4 -t- 15(- 2) = -26,^

« = |(«4-0 = f(4-26)=-176.

2. Z = 42, a = - 3, d = 3. Find n and s.

3. a = - 4, n = 8, s = 64. Find d and I.

4. d = - i, n = 6, « = 21. Find s and a.

5. d = — i, n = 10, s = 65. Find a and i.

6. s = 161, Z = 4, a = — 3. Find d and n.

7. Z = 22, s = 243, n = 13. Find a and d.

8. s = - 15, Z = - 2, d = 2. Find n and a.

9. d = 41, a = - 16, s = 140. Find n and i.

10. Insert 8 arithmetical means between 4 and 28.

11. Find expressions for n and s in terms of a, I, and d.

12. Find expressions for I and a in terms of s, n, and d.

13. Find expressions for a and s in terms of d, /, and n.

14. Find expressions for d and n in terms of s, a, and Z.

15. Find the 13th term and the sum of the series

V2-1 V2 1

2 ' 2 '2(V2-l)'""

16. Find the 10th term and the sum of the series

V3 3V3 + 2 V3 2
T"'~~6~"'"2" + 3'"-

17. Insert 4 arithmetical means between — - and

V2 2

10\/6

18. Insert 6 arithmetical means between -x - and

4

19. Insert 3 arithmetical means between and

2

20. Find the 21st term and the sum of the series , V^

V2

V2

21. Find the 10th term and the sum of the series — =

136 QUADRATICS AND BEYOND

22. Eind expressions for d and a in terms of s, Z, and n.

23. Find expressions for d and I in terms of a, n, and s.

24. Find the 8th term and the sum of the series x, 4ic, 7x, • • •.

25. Find the 9th term and the sum of the series 8, 9J, 10|, • • •.

26. Find the 12th term and the sum of the series 8, 7y\, 6|, • • • .

27. Find the 8th term and the sum of tlie series — 8, — 4, 0, • • ••

28. Find the 12th term and the sum of the series 27, 22 1, 18, • • •.

29. Find the 20th term and the sum of the series 1, — 2i, — 6, • • •.

30. Find the 11th term and the sum of the series 5, — 3, — 11, • • •.

31. Find the 9th term and the sum of tlie series x — y, x, x -]- y, • • -. *

32. Insert n — 2 arithmetical means between a and I. Write the first tliree.

Kemark. Often an exercise may be solved more simply if instead of assum-
ing the series x, x-\- y, z + 2y, ■■ -we assume x — y, x, x -\- y when three terms
are required, or x — 2y, x — y, x, x + y, x + 2y when five terms are required, or
x~-3y,x — y,x + y,x-i-3y when four terms are required.

33. The sum of the first three terms of an A. P. is 15. The sum of their
squares is 83. Find the sum of the series to ten terms.

34. Find expressions for n and a in terms of s, Z, and d. For what real
values of s, Z, and d does a series with real terms not exist ?

35. In an A. P. where a is the first term and s is the sum of the first
n terms, find the expression for the sum of the first m terms.

36. Find expressions for n and I in terms of a, s, and d. For what real
values of a, s, and d does a series with real terms not exist ?

37. If each term of the series (1), § 130, is multiplied by m, is the new
series in A. P., and if so, what are the elements of the new series ?

38. If each term of the series (1) in § 130 is increased by 6, is the new
series in A. P., and if so, what are the elements of the new series ?

39. The difference between the third and sixth terms of an A. P. is 12.
The sum of the first 10 terms is 45. Find the elements of the series.

40. Find the 10th term of an A. P. whose first and sixteenth terms are 3
and 48. Find also the sum of those eight terms of the series the last of
which is 60.

41. Two A.P.'s have the same common difference, and their first terms
are 2 and 4 respectively. The sum of the first seven terms of one is to the
sum of the first seven terms of the other as 4 is to 6. Find the elements of
both series.

42. The three digits of a number are in A. P. The number itself divided
by the sum of the digits is 48. The number formed by the same digits in
reverse order is 396 less than the original number. What is the number?

CHAPTEE XIV
GEOMETRICAL PROGRESSION

133. Definitions. A series of nTimbers such that the quotient
of any term of the series by the preceding term is always the
same is called a geometrical progression (denoted by G.P.).

The constant quotient of any term by the preceding term of a
G.P. is called the ratio.

The G.P. series 4, 8, 16, • ■ • has the ratio 2. The G.P. series 8, 4 Vi, 4, • • • has

V9

the ratio J-^i .

^' EXERCISES

Determine which of the following series are in G.P. and find the ratio.
1. 4, 2, 1, ....
3. 8, -2, .5, ....

5. Vl» i» Vl, ••••

2.

4, 8, 16, ....

4.

8, -4, -2, ....

6.

6,-21, 73i, ...

8.

^ 2 ^

V2' '' V2'

10.

V3 [3" V3
8 ' V32' 4 '

7.-^, A 2, ....

V2

\ 6 V5 Vl5

11. —J^ — -^, 5-2V6,3V3-V2,.... 12. V2 - 1, 1, V2' + 1, ....

V3 - V2

134. The nth term. The terms of a G.P. in which a is the
first term and r the ratio are as follows :

a, ar, ar^^ at^, ....

The power of r in the second term is 1, in the^ third term is
2, and in fact is always one less than the number of the term.
If we call I the nth term, we have the following expression for
the T^th term :

137 ^

138 QUADRATICS AND BEYOND

135. The sum of the series. We obtain a formula for finding
the sum of the first 7i terms of a G.P. by the following

Theorem. The sum s of the first n terms of the geometrical
progression a, ar, ar^^ ... is

a — rl .{■>
' = T^r- "
By definition, s = a -\- ar + ar^ H- • • • 4- ar"~^
= a(l -h r + r"^ -\ h r^~^)

K^)

a — rar"^ ^ a — rl

by (III), p. 15
by (I), p. 137

136. Geometrical means. The n — 2 terms between the first
and the ?ith term of a G.P. are called the geometrical means
between those terms.

If one geometrical mean is inserted between two numbers, it
is called the geometrical mean of those numbers. Thus the
geometrical mean between a and h is ^ ah.

The two fundamental formulas

« = ar»*-i, (I)

_ a{\-r^) _ a-rl

^ - 1-r - "rr7 ^^^^

contain the five elements a, I, r, n, s, any two of which may be
found if the remaining three are given.

EXERCISES

1. Find the 7th term and the sum of the G.P. 1, 4, 16, . . ..
Solution : a = 1, n = 7, r = 4.

Substituting in (I), I = ar*'-^ = 1 • 4^ = 4096,

a — rl

1-r

o V *-^ *• • /TTv 1-4-4096 1638.3 ^,^,
Substitutmg m (II), s = = = 6461.

GEOMETRICAL PROGRESSION 139

2. Insert 2 geometrical means between 4 and 32,

3. Insert 4 geometrical means between 32 and 1.

4. Insert 3 geometrical means between 3 and |f .

5. Insert 4 geometrical means between a^ and l^.

6. Insert 4 geometrical means between 1 and 9V3.

7. Insert 3 geometrical means between y- a-iid 73|.

8. What is the geometrical mean between 3 and 27 ?

9. Insert 3 geometrical means between VS and "v/24.

10. Insert 4 geometrical means between a and a^ Va6^.

11. Insert 3 geometrical means between — | and — 2^.

12. What is the geometrical mean between — 2 and — f ?

13. Find the 7th term and the sum of the series 1, 3, 9, • • •.

14. Find the 6th term and the sum of the series 2, 4, 8, • • • .

15. What is the geometrical mean between -y/a^h and VoP ?

16. Find the 7th term and the sum of the series 8, 2, .5, • • •.

17. Find the 8th term and the sum of the series ^^j, i^, j, • • • .

18. Find the 7th term and the sum of the series Vi, 2, 2V2, • • •.

19. Find the 7th term and the sum of the series \/2, V^, Vi, • • • .

20. Find the 10th term and the sum of the series ■^\-^^ y^^, ^^j, • • •.

21. Find the 5th term and the sum of the series V2 — 1, 1, 1 + V2, • • •.

22. The first and sixth terms of a G.P. are 1 and 243. Find the interme-
diate terms.

23. Find the 5th term and the sum of the series
^ :, 6-2V6, 9V3-11V2, ••..

V3+ Vii

24. Insert 3 geometrical means between — and - .

V3 9

25. What is the geometrical mean between ~ and ^^ ^?

X -\- y X -y

1 - 4

26. Find the 6th term and the sum of the series , — •\/2, — =, • • ..

V2 V2

27. Find the 6th term and the sum of the series -* /- , 1, — --, • • •.

\3 V2

28. Find the 5th term and the sum of the series v^, — 1, , • • •.

29. Find the 6th term and the sum of the series , -» / — , , • • •.

8 \32 4

140 QUADRATICS AND BEYOND

30. The geometrical mean of two numbers is 4 and their sum is 10. Find
the numbers.

31. The fourth term of a G.P. is 192, the seventh term is 12,288. Find
the first term and the ratio.

32. If the same number be added to or subtracted from each tern^^^^f^
G.P., is the resulting series geometrical?

33. The product of the first and last of four numbers in G.P. is 64.
Their quotient is also 64. Find the numbers.

34. The product of four numbers in G.P. is 81. The sum of the second
and third terms is i. Find the numbers.

35. If every term of a G.P; be multiplied by the same number m, is the
resulting series a G.P.? If so, w^hat are the elements?

36. The sum of three numbers in G.P. is 42. The difference between the
squares of the first and the second is 60. What are the numbers ?

37. The difference between two numbers is 48. The arithmetical mean
exceeds the geometrical mean by 18. Find the numbers.

38. Four numbers are in G.P. The difference between the first and the
second is 4, the difference between the third and the fourth is 36. Find the
numbers.

39. A ball falling from a height of 60 feet rebounds after each fall one
third of the last descent. What distance has it passed over when it strikes
the ground for the eighth time ?

40. The difference between the first and the last of three terms in G.P.
is four times the difference between the first and second terms. The sum of
the numbers is 208. Find the numbers.

41. An invalid on a certain day was able to take a single step of 18
inches. If he was each day to walk twice as far as on the preceding day,
how long before he can take a five-mile walk ?

42. The difference between the first and the last of four numbers in G.P.
is thirteen times the difference between the second and third terms. The
product of the second and third terms is 3. Find the numbers.

137. I nfinite series. When the number of terms of a G.P. is
unlimited it is called an infinite geometrical series.

In the series a, ar, ar^^ • • •, when r > 1, evidently each term is
larger than the preceding term. The series is then called increas-
ing. When r < 1, each term is smaller than the preceding term
and the series is called decreasing.

T,T . a (1 — r") a ar^

Now m any case s = — \ =

1 — r 1 — r 1 — r

GEOMETRICAL PROGRESSION 141

When r > 1, evidently r" becomes very large for large values
of n. For this case, then, the sum of the first n terms becomes very
large for large values of n. In fact we can take enough terms
so that s will exceed any number we may choose. If, however,
r < 1, as 71 increases in value r" becomes smaller and smaller. In
fact we can choose n large enough so that r" is as small as we wish,
or as we say, approaches as a limit. But since r" may be made
as small as we wish, ar^ also approaches as a limit, and conse-

quently approaches as a limit. Thus when r < 1 the

value of the sum of the first n terms approaches as n

1 — r

becomes very great. This we express in other words by asserting
that the sum of the infinite series

a -\- ar -\- ar^ + ■ "j when r < 1,

is s^ =

1-r

. EXERCISES

Find the sum of the following infinite series.

1. 6 + 3 + 1 + ....
Solution : a =z 6, r = |.

a

1-r

6

= 5 = 12.

i-i

h

3.

64 + 8 + 1+..

5.

^ + i^ + :rV + -

7.

2 + .5 + .125 +

2. l + i + i + ....
4. h + \ + l + -"'
6. ! + f + -V + ---

■v/2
8. V2 + l + -y- + .... 9. (V2 + l) + l + (v^-l) + ....

10. How large a value of n must one take so that the sum of the first
n terms of the following series differs from the sum to infinity by not more
than .001?

(a) 8 + 4 + 2 + ....

Solution : a = 8, r = |.

a ar'* ar**

8 = = Sao —

1-r 1-r 1

8^-8=

1-r

142

We must find for what value of n the expression

is less than .001.

say

8-2

2»

16

2«*

'"^ - - 16 1 ^

By trial we see that if w = 14 the value of — is , which is less than

.001 2". 1^2^

(b) 27 + 3 + i + . . .. (c) 4 + I + ^ij + • • •.

(d) 1 + ^1^ + 3,1^ + .... (e) 64 + 16 + 4 + ....

(f) 100 + 20 + 4 + . . •. (g) 60 + 20 + 6f% . . ..

11. What is the value of the following recurring decimal fractions?

(a) .212121....

Solution : This decimal may be written in the form

Here

(b) .333....
(e) .343343

21 21 21

100 ' (100)2 ' (100)3"'" •

21 1-
100 ' 100

5 _ « _ .21 _ .21 _ 7
* 1-r l-.Ol .99 33*

(c) .717171....

(d) .801801..

(f) 1.43131.... ^

(g) 2.61414..

CHAPTEE XV
PERMUTATIONS AND COMBINATIONS

138. Introduction. Before dealing directly with the subject of
the chapter we must answer the question, In how many distinct
ways may two successive acts be performed if the first may be
performed in p ways and the second may be performed in q ways ?
Suppose for example that I can leave a certain house by any
one of four doors, and can enter another house by any one of five
doors, in how many ways can I pass from one house to the other ?
If I leave the first house by a certain door, I have the choice of
all five doors by which to enter the second house. Since, how-
ever, I might have left the first by any one of its four doors,
there are 4 • 5 = 20 ways in which I may pass from one house
to the other. This leads to the v,^

Theorem. If a certain act may he 'performed in p ways, and
if after this act is performed a second act may he performed in q
ways, then the total numher of ways in which the two acts may
he performed is p • q.

With each of the p possible ways of performing the first act
correspond q ways of performing the second act. Thus with all
the p possible ways of performing the first act must correspond
q times as many ways of performing the second act. That is, the
two acts may be performed in ^ • 5' ways.

It is of course assumed in this theorem that the performance
of the second act is entirely independent of the way in which the
first act is performed.

143

EXERCISES

1. I have four coats and five hats. How many different combinations
coat and hat can I wear ?

Solution : The first act consists in putting on one of my coats, whicli may
be done in four ways-, the second act consists in putting on one of my
hats, which may be done in five ways. Thus I have 4 • 5 = 20 different
combinations of coat and hat.

2. In how many ways may the two children of a family be assigned to
five rooms if they each occupy a separate room ?

3. A gentleman has four coats, six vests, and eight pairs of trousers. In
how many different ways can he dress ?

4. I can sail across a lake in any one of four sailboats and row back
in any one of fifteen rowboats. In how many ways can I make the trip ?

5. Two men wish to stop at a town where there are six hotels but do not
wish quarters at the same hotel. In how many ways may they select hotels ?

6. A man is to sail for England on a steamship line that runs ten boats
on the route, and return on a line that runs only six. In how many different
ways can he make the trip ?

7. In walking from A to B one may follow any one of three roads; in
going on from B to C one has a choice of five roads. In how many different
ways can one walk from A to C ?

139. Permutations. Each different arrangement either of all
or of a part of a number of things is called a permutation.

Thus the digits 1, 2 have two possible permutations, taken both
at a time, namely, 12 and 21.

The digits 1, 2, 3 have six different permutations when two are
taken at a time, namely, 12, 13, 21, 23, 31, 32. For if we take 1
for the first place, we have a choice of 2 and 3 for the second
place, and we get 12 and 13. If 2 is in the first place, we get 21
and 23. Similarly, we get 31 and 32. In this process it is noted
that we can till the first of the two places in any one of three
ways ; the second place can be filled in each case in only two ways.
Thus by the Theorem, § 138, we should expect 3-2 = 6 permuta-
tions of three things taken two at a time. We observe that this
product 3 • 2 has as its first factor 3, which is the total number of
things considered. The number of factors is equal to the number
of digits taken at a time, i.e. two. This leads to the general

PEKMUTATIONS AND COMBINATIONS 145

Theorem. Tlie number of permutations of n ohjects taken r

at a time is , i\ , , -t\ /t\

n(n — 1) ■ ■ ' (n — r -\- J^)' v-j

This is symbolized by P„^ ^.

This formula is easily remembered if one observes that the first
factor is 7i, the total number of objects considered, and that the
number of factors is r, the number of objects taken at a time.
Thus ■Py3 = 7-6-5.

We prove this theorem by complete induction.

Flrstj let r = 1. There are evidently only n different arrange-
ment of n objects, taking one object at a time, namely (assuming
our objects to be the first 7i integers),

1, 2, 3, ..-, 71.

Let us take two objects at a time, i.e. let r= 2. Since there

are n objects, we have n ways of filling the first of the two places.

When that is tilled there are n — 1 objects left, and any one may

be used to fill the second place. Thus, by the Theorem, § 138,

there are for r = 2 . . .

n {n — V)

different permutations.

Secondy assume the form (I) for r = m^

Pn,m = n(n-l)..-(n-m + l). (1)

We can fill the first m places in P„ ^ different ways since there
are that number of permutations of n things taken m at 'a time.
This constitutes the first act (§ 138). The second act consists in
filling the m -f 1st place, which may be done in n — m ways by
using any of the remaining n — in objects. Thus the number of
permutations of n things taken 7n + 1 at a time is
Pn,r,^ + i = Pn,nr(n-m)=:=n(n-l){n-2)---(n-m + l)(n-m\
which is the form that (1) assumes on replacing m by m + 1.

Corollary. The number of permutations of n things taken all

at a time is ^ , -,\ ^ -t » * /o\

F^^^=n(n-'l)-'-2-l = n!* (^)

Taking n = r in (I), we get (2).

» « / is the symbol for 1 • 2 .3 • 4 •••(»- 1) n, and is read factorial n.

EXERCISES

1. How many permutations may be formed from 8 letters taken four at a
time?

Solution : n = 8, r=»4, n-r + l=6,

Pg, 4 = 8 • 7 . 6 . 5 = 1680.

2. In how many different orders may 6 boys stand in a row ?

3. How many different numbers less than 1000 can be formed from the
digits 1, 2, 3, 4, 5 without repetition ?

4. How many arrangements of the letters of the alphabet can be made
taking three at a time ?

5. How many numbers between 100 and 10,000 can be formed from the
digits 1, 2, 3, 4, 5, 6 without repetition ?

6. How many different permutations can be made of the letters in the
word compute taking four at a time ?

7. In a certain class there are 4 boys and 5 girls. In how many orders may
they sit provided all the boys sit on one bench and all the girls on another ?

Hint. Use Corollary § 139, and then Theorem, § 138.

8. I have 6 books with red binding and 3 with brown. In how many ways
may I arrange them on a shelf so that all the books of one color are together ?

140. Combinations. Any group of things that is independent of
the order of the constituents of the group is called a combination.

The committee of men Jones, Smith, and Jackson is the same
as the committee Jackson, Jones, and Smith. The sound made by
striking simultaneously the keys EGrC of a piano is the same as
the sound made by striking CGE. In general a question involv-
ing the number of groups of objects that may be formed where
the character of any group is unaltered by any change of order
among its constituent parts is a question in combinations.

Suppose for example that we ask how many committees of three
men can be selected from six men. If the men are called A, B, C,
D, E, F, there are, by § 139, 6 -5 • 4 = 120 difPerent arrangements
or permutations of the six men in groups of three. But the permu-
tations A, B, C ; A, C, fe ; B, A, C, etc. (3 ! = 6 in all for the men
A, B, and C), are all distinct, while evidently the six committees
consisting of A, B, and C are identical. This is true for every
distinct set of three men that we could select; that is, for the

PERMUTATIONS AND COMBINATIONS 147

six different permutations of any three men there is only one
distinct committee. Hence the number of committees is one sixth

the total number of permutations, or -^'
This leads to the general

Theorem. The number of comhinations of n things taken r at

a time is . -,^ , . ^x

n(n — I)--- (n — T-\- 1)

This is symbolized by C„^ ^.

The number of permutations of n things taken r at a time is

p^^^ = n(n - 1) • ' • (r, — r + 1).

In every group of r things which form a single combination
there are (Cor., p. 145) r ! permutations. Thus there are r ! times
as many permutations as combinations. That is,

r -?Ji^- n{n-l)-'-{n-r-\-l)
"''•""/•!" r\ ' ^^

This formula is easily remembered if one observes that there
is the same number of factors in the numerator as in the denomi-
nator. Thus

10- 9- 8
^10.3- -^.2-3
Corollary. C ^= C

n, n — r

Multiplying numerator and denominator of (I) by (n — r)!,
_ n(n-l)-'-(n-r + l)(n-r)-'-2-l
"•'•" rl(n-ry.

_ n{n-l)---{r-\-l)
(n — r)\

— ^ (^ ~ 1) • • • [^ ~ (^ ~ ^) + ^]

(n — r)\

n.n — r'

This corollary saves computation in some cases. For instance, if we wish to
compute Ci9, ir, it is more convenient to write Ci9, 17= Ci9, 2 = ^ , = 171 than the
expression for C'lo, 17.

EXERCISES

1. How many committees of 5 men can be selected from a body of 10
men three of whom can serve as chaii'man but can serve in no other capacity ?

Solution : There are 7 men who may fill 4 places on the committee.

^' 1.2.3-4

There are 3 men to select from for the remaining place of chairman,
and the selection may be made in 3 ways. Thus the committee can be
made up in 3 • 35 = 105 ways.

2. How many distinct crews of 8 men may be selected from a squad of
14 men ?

3. How many distinct triangles can be drawn having their vertices in
10 given points no three of which are in a straight line ?

4. How many distinct sounds may be produced on 9 keys of a piano by
striking 4 at a time ?

5. In how many ways can a crew of 8 men and a hockey team of 5 men
be made up from 20 men ?

6. In how many ways may the product a-b • c • d- e -f be broken up
into factors each of which contains two letters?

7. If 8 points lie in a plane but no three in a straight line, how many
straight lines can be drawn joining them in pairs?

8. How many straight lines can be drawn through n points taken in
pairs no three of which are in the same straight line ?

9. Seven boys are walking and approach a fork in the road. They
agree that 4 shall turn to the right and the remainder turn to the left. In
how many ways could they break up ?

Solution : The number of groups of 4 boys that can be formed from the

CV 4 = = o5.

4!

For each group of 4 boys there remains only a single group of 3 boys.
Thus the total number of ways in which tliQ party can divide up is
precisely 35.

10. If there are 12 points in space but no four in the same plane, how
many distinct planes can be determined by the points?

Hint. Three points determine a plane.

11. Eight gentlemen meet at a party and each wishes to shake hands
Tfith all the rest. How many hand shakes are exchanged ?

I

PERMUTATIONS AND COMBINATIONS 149

12. In how many ways can a baseball team of 9 men be selected from
14 men only two of whom can pitch but can play in no other position ?

13. How many baseball teams can be selected from 15 men only four of
whom can pitch or catch, provided these four can play in either of the two
positions but cannot play elsewhere ?

14. Two dormitories, one having 3 doors, the other having 6 doors, stand
facing each other. A path runs from each door of one to every door of the
other. How many paths are there ?

15. Show that the number of ways in which p -{- q things may be divided

into groups of p and q things respectively is ^ ^^•

p\q\

16. Out of 8 consonants and 3 vowels how many words can be formed
each containing 3 consonants and 2 vowels ?

17. A boat's crew consists of 8 men, three of whom can row only on one
side and two only on the other. In how many ways can the crew be arranged ?

18. A pack of cards contains 62 distinct cards. In how many different
ways can it be divided into 4 hands of 13 cards each ?

19. Five points lie in a plane, but no three in any other plane. How
many tetrahedrons can be formed with these points taken with two points
not in the plane ?

141. Circular permutations. By circular permutations we
mean the various arrangements of a group of things around a
circle.

Theorem. The number of orders in which n things may he
arranged in a circle is (n—l) !.

Suppose A is at the point at which we begin to arrange the
digits 1, 2, 3, • • •, n. Suppose we start our arrangement of digits
at A with a given digit a. We have then
virtually n—l places to fill by the remaining
n—l digits. Thus we get {n—l)\ (p. 145)
permutations of the n digits keeping a fixed.
But suppose we start our arrangement, that is,
fill the place at A with any other digit, as h,
and the remaining places in any order what-
ever. If we now go around the circle till we
come to the digit a, the succession of digits from that point
around the circle to a again must be one of the {n —1)\ orders

which we obtained when we took a as the initial figure. Thus
the only distinct orders in which the n digits can be arranged
on a circle are the (n —1)\ permutations we obtained by filling
the first place with a.

EXERCISES

1. In how many orders can 6 men sit around a circular table ?
Solution :

n = Q, n - 1 = 5, (n - 1)! = 5 ! = 120.

2. In how many ways can 8 men sit around a circular table ?

3. In how many ways may the letters of live be arranged on a circle ?

4. In how many ways may the letters of permutation be arranged on a
circle ?

5. In how many ways can 4 men and 4 ladies sit around a table so that
a lady is always between two men ?

6. In how many ways may 4 men and their wives be seated around a table
so that no man sits next his wife but the men and the women sit alternately ?

7. In how many ways can six men and their wives be seated around a
table so that each man sits between his wife and another lady ?

8. In how many ways can 10 red flowers and 5 white ones be planted
around a circular plot so that two and only two red ones are adjacent ?

142. Theorem. The number of permutations of n things of

. n!
which p are alike, taken all together y is -^•

If all the things were different, we should have n ! permutations.
But since p of the n things are alike, any rearrangement of those
p like things will not change the permutation. Eor any fixed
arrangement of the n things there are p ! different arrangements

of the p like things. Thus — : of the n\ permutations are iden-

tical, and there are only — '- distinct permutations of the n things
^p of which are alike. '

Corollary. If of n things p are of one kind, q of another

n f
kind, r of another, etc., then there are — ; — -^— — permutations

of the n things taken all at a time. -^ ' ^' ' ^

PERMUTATIONS AND COMBINATIONS 151

EXERCISES

1. How many distinct arrangements of the letters of the word Cincinnati
are possible ?

Solution : There are in all 10 letters, of which 3 are i, 2 are c, and 3 are n.
Thus the number of arrangements is

10! _ X.2.^-^-5-^-7-8-9.10
3!3!2!" ;.^.3X-^-^;-;Z
= 2.5-7.8.9.10 = 50,400.

2. How many distinct arrangements of the letters of the word parallel
can be formed ?

3. How many signals can be made by hanging 15 flags on a staff if 2
flags are white, 3 black, 5 blue, and the rest red ?

4. How many signals can be made by the flags in exercise 3 if a white
one is at each extreme ?

5. How many signals can be made by the flags in exercise 3 if a red flag
is always at the top ?

6. Would 3 dots, 2 dashes, and 1 pause be enough telegraphic symbols
for the letters of the English alphabet, the numerals, and six punctuation
marks ?

CHAPTEE XVI
COMPLEX NUMBERS

143. The imaginary unit. When we approached the solution
of quadratic equations (p. 52) we saw that the equation x^ = 2
was not solvable if we were at liberty to use only rational num-
bers, but that we must introduce an entirely new kind of number,
defined as a sequence of rational numbers, if we wished to solve
this equation. The excuse for introducing such numbers was not
that we needed them as a means for more accurate measurement,
— the rational numbers are entirely adequate for all mechanical
purposes^ — but that they are a mathematical necessity if we
propose to solve equations of the type given.

A similar situation demands the introduction of still other
numbers. If we seek the solution of

£C2=-1,

(1)

we observe that there is no rational number whose square is — 1.
Neither can we define V— 1 as a sequence of rational numbers
which approach it as a limit. We may write the symbol V— 1, but
its meaning must be somewhat remote from that of V2, for in
the latter case we have a process by which we can extract the
square root and get a number whose square is as nearly equal to
2 as we desire. This is not possible in the case of V^l. In fact
this symbol differs from 1 or any real number not merely in
degree but in kind. One cannot say V— 1 is greater or less than
a real number, any more than one can compare the magnitude of
a quart and an inch.

V — 1 is symbolized by I and is called the imaginary unit. The
term "imaginary" is perhaps too firmly established in mathe-
matical literature to warrant its discontinuance. It should be
kept in mind, however, that it is really no more and no less

152

COMPLEX NUMBERS 153

imaginary than the negative numbers or the irrational numbers
are. So far as we have yet gone it is merely that which satis-
fies equation (1). When, however, we have defined the various
operations on it and ascribed to it the various characteristic
properties of numbers we shall be justified in calling it a
number.

Just as we built up from the unit 1 a system of real numbers,
so we build up from V— 1 = i a system of imaginary numbers.
The fact that we cannot measure V— 1 on a rule should cause
no more confusion than our inability exactly to measure -y/2 on a
rule. Just as we were able to deal with irrational numbers as
readily as with integers when we had defined what we meant
by the four operations on them, so will the imaginaries become
indeed numbers with which we can work when we have defined
the corresponding operations on them.

144. Addition and subtraction of imaginary numbers. We
write

= 0i,

i -\- i = 2 ij
i -{- i-\ -{- i = ni. (I)

Also just as we pass from a rational to an irrational multiple
of unity by sequences, so we pass from a rational to an irrational
multiple of the imaginary unit. Thus we write a V— 1, or ai,
where a represents any real number. Consistently with § 76 we
write

^ V-a2 = ^ Va2.(_l) = ± V^ . V^ = ±a V^ = ± ai. (II)

We speak of a positive or a negative imaginary according as
the radical sign is preceded by a positive or a negative sign.
We also define addition and subtraction of imaginaries as

follows :

ai ± bi = (a ±b) i, (III)

where a and b are any real numbers.

Assumption. The commutative and associative laws of multi-
plication and addition of real numbers, § 10, we assume to hold
for imaginary numbers.

145. Multiplication and division of imaginaries. We have
already virtually defined the multiplication of imaginaries by
real numbers by formula (I). Consistently with § 76 we define

V^ . V^ = ii = i^ = - 1.

Thus V— a • V— b = Va • V^ i i = -\/ab • (— 1) = — -Vab.

The law of signs in multiplication may be expressed verbally
as follows :

The product of imar/inaries with like signs before the radical
is a negative real number. The product of imaginaries with
unlike signs is a positive real number.

For instance, - yP^ - V^^ = - 2 • 3 • i2 = 6.

We also note that

i2 = — 1, i3 = — i, i* = 1, i^ =r i, . . . ,
And, in general, Hn + k _ ik^ /c = 0, 1, 2, 3.

We define division of imaginaries as follows :

/ I r Vo^ • i la

■Vb-i ^b

In o perating with imaginary numbers, a number of the form
V— a should always be written in the form Va i before per-
forming the operation. This avoids temptation to the following

error:

V- a • V- b = V(- a) ■(-b) = -slab.

EXERCISES

Simplify the following :

1. V^8 ■ yT^.

Solution : V^^ • V^^ = VS • i . V2 • i = V2 • 8 . {2 = 4 . (_ 1) = _ 4.

2.1.

1 2^ /^ ■^— %

Solution : — = - = = = — i.

i^ i8 (i4)2 \

COMPLEX NUMBEKS

165

3.

i".

6.

V-36.

9.

V- X2«.

12.

V2 V- 8.

15.

1

18.

V-6

4. 124.

5. ii3.

7. V-64.

8. 2i'Si.

10. V-Sx^a^.

11. V-x2.

13. V-2V-6.

14. V-3%

16.1.

17.. f^.

V-2

19. V-i2.

20. V-i^.

146. Complex numbers. The solution of the quadratic equation
with negative discriminant (p. 71) affords us an expression which
consists of a real number connected with an imaginary number
by a H- or — sign. Such an expression is called a complex number.
It consists of two parts which are of different kinds, the real
part and the imaginary part. Thus 6 + 4 z means G I's + 4 ^'s.
Obviously, to any pair of real numbers (x, y) corresponds a complex
number x + ty, and conversely.

147. Graphical representation of complex numbers. We have
represented all real numbers on a single straight line. When we
wished to represent two numbers simultaneously, we made use of
the plane, and assumed a one-to-one correspondence between the
points on the plane and the pairs of numbers (cc, y). The general
complex number x -f- iy depends
on the values of the independent
real numbers x and y, and may
then properly be represented by
a point on a plane. We repre-
sent real numbers on the X axis,
imaginary numbers on the Y axis,
and the complex number x + iy
by the point (x, y) on the plane.
Thus the complex numbers 6 + ^ 3,
— 4 -f 1 4, 7 — i 5, — 2 — 1 4 are represented by points on the plane
as indicated in the figure.

148. Equality of complex numbers. We define the two com-
plex numbers a -f ib and c -{- id to be equal when and only when
a = c and h — d.

y>

k

-

4+

i4

6f

i?,

X

-2-

i4

7-

15

rn

Symbolically a + ii ^ c + id

when and only when a = c, b = d.

The definition seems reasonable, since 1 and i are different in
kind, and we should not expect any real multiple of one to cancel
any real multiple of the other.

Similarly, if we took not abstract expressions as 1 and i for
units but concrete objects as trees and streets, we should say that

a trees + b streets = c trees + d streets
when and only when a = c and b = d.

Principle. When two numerical expressions involving imagi-
naries are equal to each other, we may equate real parts and
imaginary parts separately.

The graphical interpretation of the definition of equality is that equal complex
numbers are always represented by the same point on the plane.

From the definition given we see that a -\-ib = when and
only when a = b = 0.

Assumption. We assume that complex numbers obey the com-
mutative and associative laws and the distributive law given in
§ 10. We also assume the sa.me rules for parentheses as given
in § 15.

This assumption enables us to define the fundamental opera-
tions on complex numbers.

149. Addition and subtraction. By applying the assumptions
just made we obtain the following symbolical expression for the
operations of addition and subtraction of any two complex num-
bers a -{-ib and c -{- id:

a -{• ib ± (c -\- id) = a ± c -{- i(b ± d).

Rule. To add (subtract) complex numbers, add (subtract) the
real and imaginary parts separately.

150. Graphical representation of addition. We now proceed
to give the graphical interpretation of the operations of addition
and subtraction.

COMPLEX NUMBERS

167

Theorem. The sum of two numbers A = a-\- ib and B = c -\- id
is represented hy the fourth vertex of the parallelogram formed
on OA and OB as sides.

Let OASB'hQ ?i parallelogram. Draw
ES _L OE, AH A. ES, BD (= d) _L OE.
A AHS = A ODE since their sides are
parallel, and OB — AS.

Thus

Thus

DB =

: HS = d,

0D =

AH==c.

ES =

EH + HS

Y

1

d A

7

S
H

b

D

—a, >

JF E X

b + d,

OE = OF + FE = a -\- c,

and S has coordinates (a -\- c, b -{- d) and represents the sum of
A and B, by § 149.

EXERCISES

1. The difference A — B of two numbers A = a + ib and B = c + id is
represented by the extremity D of the line OD drawn from the origin par-
allel to the diagonal BA of the parallelogram formed on OB and OA as

2. Represent graphically the following expressions.

(a) 1 + i (b) - 4 - 2 i ■

(c) 6 - i. (d) - 8 + 4 i.

(e) 2 + 4 I (f) (1 + i) + (2 + i)'

(g) {2-i)-{6-Si). (h) (l_i)-(l_2i).

(i) (2 + 4i)-(l-3i). (j) 4(l + i)-2(2-3i).

(k) (6-2i) + (2 + 3i). (1) (5 + 3i) + (-l-6i).

151. Multiplication of complex numbers. The assumption of
§ 148 enables us to multiply complex numbers by the following

Rule. To multiply the complex number a -\- ib by c ■\- id, pro-
ceed as if they were real binomials, keeping in mind the laws for
multiplying imaginaries.

Thus a -\-ib

c -f- id

ac + icb -f- iad -\- (i)* bd = ac — bd + i(cb -f- ad).

152. Conjugate complex numbers. Complex numbers that differ
only in the sign of their imaginary parts are called conjugate com-
plex numbers, or conjugate imaginaries.

Theorem. The sum and the product of conjugate complex
numbers are real numhers.

Thus a -\- lb -\- a — ib = 2 a,

(a + ib) (a - ib) =a^ + b\

153. Division of complex numbers. The quotient of two com-
plex numbers may now be expressed as a single complex number.

EuLE. To express the quotient — in the form x + ^y,

rationalize the denominator^ using as a rationalizing factor
the conjugate of the denominator. ^_^

™,, a -{- ib a -\- ib c — id

Thus — = ~ ^^—^

c -\- id c -\- id c — id

_ ac -\- bd — i {ad — be)

~ c' + d^

ac -{- bd .ad — be

^ > + d' ~ "" VT^' ^^

We have now defined the fundamental operations on complex
numbers and shall make frequent use of them. If the question
remains in one's mind, "After all, what are they? " the answer is
this : " They are quantities for which we have defined the funda-
mental operations of numbers and, since they have the properties
of numbers, must be called numbers, just as a flower that has all
the characteristic properties of a known species is thereby deter-
mined to belong to that species." Furthermore, our operations
have been so defined that if the imaginary parts of the complex
numbers vanish and the numbers become real, the expression
defining any operation on complex numbers reduces to one defin-
ing the same operation on the real part of the number. Thus in
(1) above, if b = d = 0^ the expression reduces to

a _a

a c

COMPLEX NUMBERS 159

EXERCISES

Carry out tlie indicated operations.

1. (2 + V"r2)(4+V^^).
Solution : 2 + V^^ = 2 + ■v/2(-l) = 2 + z V2
4 + V:^ =:4+V5(-l) = 4 + t V6

8 - VlO + i4V2 + i2\/5

2. 5 - V2 - i Vs.

Solution :

5^ ^ 5(V2 + zV3) ^ 5V2 + Z5V3 _ /g + i Vs

\/2-iV3 (V2-iV3)(V2 + iV3)~ 2 + 3

3. (l + i)l 4. (l + i)3.
Hint. Develop by the binomial theorem.

5. {a + ib)K 6. (V^ + v^ir^)^

V7. (x + %)2 8. (x + i2/)2 + (X - %)2.

9. vT+l.Vnri. 10. (V3 + iv^)(V2 + iV3).

11. (VTT~i + Vr^y. 12. (aV6 + icVd)(aV6-icVd).

13. (Va + i V6) ( Va - i Vb). 14. (2 V7 + i3 Vs) (3 V7 - ilOV2),

15. i±i^. 16. l±i. 17. ^

l-iV3 1-i V2+V-1

18. ^ . 19. 11^. 20. (^l±i^)'.

l + V-3 (l + i)3 \ 2 /

2„ g + i Vl - a2
a — i Vl — a^

24. ;^-^- 25. -^1— _. 26. ^ + ^p^ .

27. ^^ + ^^'^ 28 (l±i^y. 29. ^ + ^^ _j_ c + id

V3-iV2 \2/ a-ib c -id

30. ?? 31. — ?i=. 32. -V+ 1

4 + 7V^6 i + 3v^^. (1 + 0' a-*y

33 VI + « + ^ Vl- <^ _ Vl - g + i Vl + g
Vi -j- g — i VT — g Vl — a — i Vl + a

34. Find three roots of the equation x^ — 1 = and represent the roots
as points on the plane.

35. Find four roots of the equation x* — 1 = and represent the roots as
points on the plane.

36. Find six roots of x^ — 1 = and represent the roots as points on the
plane. Show graphically that the sum of the six roots is zero.

37. Find three roots of x^ — 8 = and represent the roots as points on
the plane. Show graphically that the sum of the three roots is zero.

154. Polar representation. The graphical representation of
complex numbers given in § 147 gives a simple graphical inter-
pretation of the operations of addition and subtraction, but the
graphical meaning of the operations of multiplication and divi-
sion may be given more clearly in another manner. We have
seen that we may represent x -f iy by the point P (x, y) on the
plane, Represent the angle between OP and the X axis by/^.
This angle is called the argument of the complex number x -f iy.

Eepresent the line OP by p. This is called
^ , . , the modulus of ic -f iy. Then from the figure

a; = p COS B, (1)

— ^ y = psmO, (2)

x' + f = p\ (3)

Hence the complex number x + iy may be written in the form

X -\- iy = p (cos -}- i sin 0)j (4)

■when the relations between x, y and p, are given by (1), (2), and
(3). A number expressed in this way is in polar form, and may
be designated by (p, 6). We observe that a complex number
lies on a circle whose center is the origin and whose radius
is the modulus of the number. The argument is the angle
between the axis of real numbers and the line representing the
modulus.

155. Multiplication in polar form. If we have two numbers
p (cos 6 -\-i sin^) and /o'(cos 6' -\- i sin ^'), we may multiply them
and obtain

COMPLEX NUMBERS 161

p (cos 6 + isin 0)p'(Gos 0' + i sin 6')

= pp' [(cos cos 0' — sin 6 sin d')
+ i (sin \$ cos d' + cos sin ^')]
By the addition theorem ,r //i , /if\ , • • /h , /if\n /-i\

in Trigonometry = PP C^^^ (^ + ^') + ^ sm (d + d')] (1)

= R (cos © + * sin ©) . (2)

In this product pp' is the new modulus and 0-^0' the new
argument. We may now make the following statement: The
product of the two numbers p (cos 6 -\- i sin 6) and p'(cos 6' + * sin 0')
has as its modulus pp' and as its argument + 0'. Thus the
product of two numbers is represented on a circle whose radius
is the product of the radii of the circles on which the factors are
represented. The argument of the product is the sum of the
arguments of the factors.

156. Powers of numbers in polar form. When the two factors

of the preceding section (p, 0) and (p', 6') are equal, that is,

when p = p' and 6 = 6', the expression (1) assumes the form

[p (cos e + i sin 0)^ = p^ (cos 2 ^ + t sin 2 0). (1)

This suggests as a form for the nth. power of a complex number
[p (cos d -\-i sin \$)']" = p« (cos nO -\- i sin nO). (2)

The student should establish this expression by the method of
complete induction. The theorem expressed by (2) is known as
DeMoivre's theorem. Stated verbally it is as follows : The modulus
of the nth. power of a number is the nth power of its modulus. The
argument of the nth power of a number is n times its argument.

EXERCISES

Plot, find the arguments and moduli of the following numbers and of

eir products.
1. 1 + iVS, V3 + i.
Solution :

Let ^/S -\- i=p (cos ^ + i sin 5),
1 4- i V3 = /(cos d' + i sin 6^.

2.

30°;

Y.

B

Ay ! i 1 !

Then by (1), (2), (3), § 154, p = 2 ; p' =
1 = 2 sin d, hence 6 =

'--- -1— J 1 X

1 = 2 cos d", hence 6' =

60°.

= 30°, ^' = 60°

Thus if the product has the form R{cos@ + isin©), we have by § 1(
R = ppr = 4, © = ^ + ^' = 90°.

2. l+i,2 + i.

3. (l-i)3.

4. 3 + 3t, 2^iVl2.

5. 2i, l-iV3.

'4*'-^)-

7.-1 + 1,-2-2.

o , . V2 iV2
^- '^'2 2 •

'-2+ 2 ' 2 + 2

10. [2 (cos 15° + i sin 15°)]8. 11. [i (cos 30° + i sin 30°)]4.

12. [| (cos 120° + i sin 120°)]2. 13. [2 (cos 135° + i sin 135°)]*.

14. [f (cos 180° + isin 180°)] 3. 15. [f (cos 315°+ isin 315°)]2.

157. Division in polar form. If we liave, as before, two com-
plex numbers in polar form (p, 6) and (p', \$'), we may obtain their
quotient as follows.

p (cos -{- i sin \$)
p' (cos 0' -^ isin 0')

_ pp' (cos -{- i sin 0) (cos 0' — i sin 0')
~ p'2 (cos ^' + 2 sin \$') (cos ^' — i sin ^')
^ pp' [cos (6 - ^0 + i sin (^ - ^')]
~ p'2(cos2^' -f sin^^')

Rationalizing,

§ 152 and § 153,

Since sin2 + cos2 6=1, = -^ [cos (0 — \$') + i sin (^ — 0')']

= i2 (cos^Vf i sin ©).

We may now make the following statement: The, quotient of
two complex numbers has as its modulus the quotient of the moduli
of the factors, and as its argument the difference of the arguments
of t/ie factors.

158. Roots of complex numbers. We have seen that the square
of a number has as jts modulus the square of the original modulus,
while' the argument is twice the original argument.

Thjs would suggest th^-t the square root of any number, as (p, 0),

' : ' - 9

would have Vp as its modulus and - as its argument. Since

'^ ^

every yeal number has two square roots, we should expect the

same fact to hold liere. Consider the two numbers

COMPLEX NUMBERS

163

Vp /^cos I + i sin I j and Vp cos (^ -f 180° j + i sin ( | + 180° j j ,

where Vp is the principal square root of p(§ 72). The square of
the first is (p, ^), by § 155. That the square of the second is the
same is evident if we keep in mind the fact that

cos (0 + 360°) = cos e
and sin (^ + 360°) = sin a

Thus V/> (cos e + i sin 6)

Vp ( cos - + i SlUv-

or

Vp

+ 180° -f j sin/ ^ + 180

■)]■

The graphs of these two numbers are situated at points sym-
metrical to each other with respect to the origin.

We may obtain as the corresponding expression for the higher
roots of complex numbers the following :

■v^/>(cos^ + ?:sin^)= V^ I (

(9 + /c360°\ . . /^ + A;360'

cos I + I sin

j_ \ 71 / \ n

where for a given value of /i, k takes on the values 0, 1, • • •, t^ — 1,
and where VP indicates the real positive 7ith root of p.

EXERCISES

Perform the indicated operations and plot
1. 2-2V3i--l+i.
Solution :

Let \-\-i — p (cos + i sin ^),

2 -- 2 V3 i = /(cos e'-\- i sin 6").
Then p = \/12 + 1- = V2,

/ = V22 + (-2V3)^ = 4.

By (1) and (2), § 154,

sin ^ = cos ^ = — ^ , hence 6 = 46°.
V2

'4;

'^-^

(a-iaW;

164

Similarly,

300°

sin^ =

-

2V3
4

V3
2

cos^ =

f

= 1, hence d' -

2 _ 2 V3 i 4 (cos 300° + i sin 300°)

Thus = i2(cos© + isin©)=— 7^- -— . . ,^^ ■

1+i ^ ' V2(cos46°4- tsin45°)

Hence by § 157, iJ = -^ = 2v^, © = 300° - 45° = 255»

V2

/

2. V-2 + 2V3i
Let -2 + 2 V3i = p(cos^4'isin^).
Then (§ 154) /> = 4, cos^= - f = -^,

-2+i2A^

and

120^=

V-2 + 2V3i = V4 (cos 120° + i sin 120°)

T5 «i^Q /if /120° + A:360°\
By §158, =V4|cos( ^ j

l+iVs"

. . /120° + A;360°\']
.sm( ^)J

+ ^
(where A; = or 1)

= 2 (cos 60° + i sin 60°) = 1 + i V3, when k-0.
= 2 (cos 240° + i sin 240°) = - 1 - i V3, when A; = I

. 3. VV2 + i V2.

4. V

1 + i -^ 1 - i.
1 V3

6.

1 iV3

2 2

^1 + i.

8 -l-i-^-^l + l N

2 2 ■ 4 4*

10. -2\/2-2 V2i-T--2 + 2V3i

7. i + i^:t_j:Jl
2 2

9. 2-iVl2-4-3 + 3i.

11. \^l(cosl5°+isinl5°).
Solution :

"/TT ^:,o . - - -. ^Q^ '/tF /15° + fe • 360°\ . . . /16°+ fc • 360°\ 1

Vl(cosl5°+ ism 15°) = VI cosi ■ ) + ism I 1

(where fc = 0, 1, or 2).

1 (cos 5° + t sin 5°), when A; = 0,
1 (cos 125° + i sin 125°), when A; = 1,
1 (cos 245° + i sin 245°), when A; = 2.

12. Yi.

13. ^^161.

14. •V'2-f- 2\/3i.

15. v'cos330°+ tsin330<».

COMPLEX KUMBERS

165

16. V27(cos75°+ism75°). 17. Vl6 (cos 200° + i sin 200°).

18. Solve the following equations and plot their roots.

(a) x5 - 1 = 0.

Solution : x^ = 1, or a; = VT.

Let 1 = 1 + • i = p (cos ^ + i sin 6). Then p = l, 6 = 0°.

X = V 1 (cos 0° 4- i sin

0°) = Vircos/*

0° + A; . 360'

(where k takes on the values 0, 1, 2, 3, 4)

f cos 0° + i sin 0° = 1, when k = 0,
cos 72° + i sin 72°, when A; = 1,
cos 144° + i sin 144°, when fe = 2, •
cos 216° + t sin 216°, when A; = 8,
cos 288° + i sin 288°, when A; = 4.

These numbers we observe lie on a circle of
unit radius at the vertices of a regular pentagon.

(b) a;4 - 1 = 0.
(e) x6 - 1 = 0.

J

(C) X3-1 = 0.
(f ) X8 - 1 = 0.

C?x~^ <Jtf^.C<.

(d) x5 - 32 = 0.
(g) x8 -- 27 = 0.

J^l^^

y.

O^

CHAPTEE XVII
THEORY OF EQUATIONS

159. Equation of the nth degree. Any equation in one variable
in whicli the coefficients are rational numbers can be put in the

form

f (x) =aoX'^ + a^x--' -}--'• + a, = 0, (1)

where Gq is positive and a^, • • •, a„ are all integers.

The symbol /(x) is read "foix" and is merely an abbreviation for the right-
hand member of the equation. Often we wish to replace x in the equation by
some constant, as a, — 2, or 0. We may symbolize the result of this substitution
by/(a),/(-2),or/(0).

Thus f{b) = a(fi» + aift"" ^ + ^ • • + a„ .

We symbolize other expressions similarly by <f) (x), Q{x), etc.

When we speak of an equation we assume that it is in the form
of (1). This equation is also written in the form

a;" + M"-i + ---^„ = 0, (2)

where bi = —> b^ = —> •• • K — ~'

The 5's are integers only when ai, «2j • • -, «„ are multiples of Qq.

160. Remainder theorem. We now prove the following impor-
tant fact.

Theorem. When f(x) is divided hy x — c, the remainder is
f(x) with c substituted in place of the variable.

Divide the equation (1) by x — c. Let R be the remainder,
which must (§ 26) be of lower degree in x than the divisor ;
that is, in this case, since x — c is the divisor, R must be a con-
stant and not involve x at all. Let the quotient, which is of
degree n ~ 1 in x, he represented by Q (x).

160

THEORY OF EQUATIONS 167

Then i^ = Q(x)+ -5—

X — C ^ ^ X — G

Clearing of fractions,

f(x)= Q(x){x-g)+R.

But since this equation is an identity it is ahvays satisfied
whatever numerical value x may have (§ 53).

Let X = c.

Then /(c) = a,c- + %c"-i + ... + «„= Q (c) (c - c) + i2.

But since c — c = 0, Q(c) (c — c) = 0, and

R = aoc" + ^ic"- 1 H h a„ =/(c)-

Corollary. If c is a root of f(x) = 0, then x — c is a factor
of the left-hand member. ,

For if c is a root of the left-hand member, it satisfies that
member and reduces it to zero when substituted for x. Thus by
the previous theorem we have, since

aoC* H- o^ic"-^ H a^ = R = 0,

f{x)=Q(x)(x-c).

161. Synthetic division. In order to plot by the method of
§ 103 the equation

y = a^x"" + a^x""- 1 H \- a^,

when the a's are replaced by integers, we should be obliged
laboriously to substitute for x successive integers and find corre-
sponding values of y, which for large values of n involves con-
siderable computation. We can make use of the preceding theorem
to lighten this labor. The object is to find, with the least possible
computation, the remainder when the polynomial f(x) is divided
by a factor of form x — c, which by the preceding theorem is the
value of f(x) when x is replaced by c, that is, the value of y
corresponding to x = c. For illustration, let

f(x) = 2 x^ - 3 a;« -h ?c2 - a: - 9 and c = 2.

By long division we have

a;-2| 2x^-3a;^H- ^' - x- 9 |2a;« + o^' + 3a; + 5

1 x» + a;2
lx3-2a;2

3a;2- X

3a;2-6x

5a;- 9
5a; -10

+ 1

We can abbreviate this process by observing the following
facts. Since x is here only the carrier of the coefficient, we may
omit writing it. Also we need not rewrite the first number of
the partial product, as it is only a repetition of the number
directly above it in full-faced type. Our process now assumes
the form

1- 21 2-3 + 1-1- 9 |2 + l + 3 + 5

:l-4
+ 1 -? /

/ -2

+ 3 ,'
-6

+ 5

10

+ 1

Since the minus sign of the 2 changes every sign in the partial
product, if we replace — 2 by + 2 we may add the partial prod-
uct to the number in the dividend instead of subtracting. This
is also desirable since the number which we are substituting for
X is 2, not — 2. Thus, bringing all our figures on one line and
placing the number substituted for x at the right hand, we have

2-3 + 1-1- 9[2

+ 4 + 2 + 6 + 10
2+1+3+5+ 1

THEORY OF EQUATIONS 169

We observe that the figures in the lower line, 2, 1, 3, 5, up to
the remainder are the coefficients of the quotient 2 x^ -{- x^ -\- S x -\- 5.

EuLE FOR SYNTHETIC DIVISION. Write the coefficients of the
polynomial in order, supplying when a coefficient is lacking.

Multiply the number to he substituted for x by the first coeffi-
cient, and add (algebraically) the product to the next coefficient

Multiply this sum by the number to be substituted for x, add to
the next coefficient, and proceed until all the coefficients are used.
The last sum obtained is the remainder and also the value of
the polynomial when the number is substituted for the variable.

162. Proof of the rule for synthetic division. This rule we now
prove in general by complete induction. Let the polynomial be

a^x"" 4- a^x^-^ + a^x""-^ -\ h «„.

Let the number to be substituted for x be a.

First. Let n = 2. Carry out the rule on a^x"^ + a^x + a^.

We have , , .

+ aQ(X + {apa + a\) cc

<^oi + «o«^ + «i, + {a^a -|- a.i) a + ^2 = a^a^ + a^a, + a^.

Second. Assume the validity of the rule for n = m, and prove

that its validity f or ti = m + 1 follows. Assume then that the rule

carried out on

f{x) = a^x^ + a^x^-^ + ••• + «„

affords the remainder

a^a"^ + ai^"*-! ^ -\- a^ =f((x)'

Now the polynomial of order w + 1 is

^0^"'+^ 4- aix"" H \-a^x-i- a^ + j = x •f(x) + a^^^

Hence the next to the last remainder obtained by applying the
rule to this polynomial would be f(cc), since the succession of
coefficients is the same for both polynomials up to a^+i- By
the rule the final remainder is obtained by multiplying the expres-
sion just obtained, in this case f(cc), by a and adding the last
coefficient, in this case a^_^,^. This affords the final remainder

170

EXERCISES

1. Prove by complete induction that the partial remainders up to the final
remainder obtained in the process of synthetic division are the coefficients of
the quotient of f{x) hj z — a.

2. Perform by synthetic division the following divisions.

(a) a;3 - 7x2 _ 6x + 72 by a; - 4.

Solution : 1-7-6 + 72 [4

4 _ 12 - 72

1 _ 8 - 18
Quotient = a;2_3x- 18.

(b) aj8 - 9x + 10 by X - 2. (c) 4x3 - 7 x - 87 by x - 3.

(d) x3 + 8x2 - 4x - 32 i3y x-2. (e) x^ + 4x2 - 7x - 30 by x + 3.

(f) x8 - 6x2 + 11 X - 6 by X - 1.

(g) x* - 16x3 + 86x2 - 176x + 105 by x2 - 8x + 7.

Hint. SinceK2— 8a;+7=(a; — 7)(x — 1), divide by x—7 and the quotient by a;— 1.

(h) x6 + 1 by X + 1. (i) x9 - 1 by X - 1.

(j) x4 + x3 - X - 1 by x2 - 1. (k) x^ - 2x3 - 4x by x - 3.

(1) x6 - 2x8 - 4x - 1 by X + 2. (m) 4x3 - 6x2 - 2x - 1 by x - 3.

(n) 2x* + 5x3 - 37x2 +44x + 84 by x2 + 6x - 6.

163. Plotting of equations. We can now form the table of
values necessary to plot an equation of the type

^0^" + ^i^'"" ^ H 1- «„-i^ + «„ = y.

Example. Plot x3 + 4 x2 - 4 = j/.
l + 4 + 0-4[l

+1+5+5
1+6+6+1
1 + 4 + 0-41-1

-1-3+3
l+S-S-l

1 + 4 + 0-41-2

-2-4+8
1+2-4+4
l_f.4 + 0-4[-_3

-3-3+9

+1-3+5
1 + 4 + 0-41-4

-4+0+0

X

V

X

y

-4

- 1

-1

1

+ 1

-2

+ 4

;l xo

-3

+ 6

-4

-4

1+0+0-4

THEORY OF EQUATIONS 171

In this figure two squares are taken to represent one unit of x. A single
square represents a unit of y.

By an inspection of the figure it appears that the curve crosses the X axis
at about x = .8, x = — 1.2, and x = — 3.7. Thus the equation for y = has
approximately these values for roots (§ 110).

164. Extent of the table of values. Since the object of plot-
ting a curve is to obtain information regarding the roots of its
equation, stretches of the curve beyond all crossings of the X axis
are of no interest for the present purpose. Hence it is desirable
to know when a table of values has been formed extensive enough
to afford a plot which includes all the real roots. If for all values
of X greater than a certain number the curve lies wholly above
the axis, there are no real roots greater than that value of x.

By inspection of the preceding example it appears that if for
a given value of x the signs of the partial remainders are all
positive, thus affording a positive value of ?/, any greater value
of X will afford only positive partial remainders and hence only
positive values of y.

Thus when ■ all the partial rem.ainders are positive no greater
positive value of x need he substituted.

Similarly, when the partial remainders alternate in sign begin-
ning with the coefficient of the highest power of x, no value of x,
greater negatively, need be substituted.

In plotting, if the table of values consists of values that are
large or are so distributed that the plot would not be well propor-
tioned if one space on the paper were taken for each unit, a scale
should be so chosen that the plot will be of good proportion,
that is, so that all the portions of the curve between the extreme
roots shall appear on the paper, and the curvatures shall not be
too abrupt to form a graceful curve. This was done, for example,
in the figure, § 163.

EXERCISES

Plot and measure the values of the real roots of the equations when y = 0.
1. x8 - 7 X - 6 = y. 2. x3 - 7 X + 5 = y.

3. 7x8 - 9x - 6 = y. 4. x3 - 31 X + 19 = y.

6. x3 - 12x - 14 = y. 6. 4x8 - 13x + 6 = y.

172 abvakced algi:biia

7. a;8 - 12x - 16 = y. 8. x^ - 45x + 152 = y.

9. x* - 2x3 - X + 2 = y. 10. 8x3 _ igxS + 17x - 6 = y.

11. X* - 17x2 + X + 20 = y. 12. x-i - 4x3 + 9x2 - 8x + 14 = y.

13. 18x3-36x2 + 9x + 8 = 2/. 14. x* + 5x3 + 12x2 + 52x- 40 =y.

15. x*-2x3-7x2+19x-10=y. 16. x4-6x3 + 3x2 + 26x-24 = y.
17. 6x4 - 13x3 + 20x2 - 37x + 24 = y.

165. Roots of an equation. In the case of the linear and
quadratic equations we have been able to find an explicit value
of the roots in terms of the coefficients. Such processes are prac-
tically impossible in the case of most equations of higher degree.
In fact the proof that any equation possesses a root lies beyond
the scope of this book, and we make the

Assumption. Every equation possesses at least one root.

This is equivalent to the assumption that there is a number,
rational, irrational, or complex, which satisfies any equation.

166. Number of roots. We determine the exact number of
roots by the following

Theorem. Every equation of degree n has n roots.

Given the equation /(x) = ao^?" + aiic""^ -\ +- a„ = 0.

Let ^i (see assumption) be a root of this equation. Then (p. 166)
ic — ^i is a factor of the left-hand member, and the quotient of
f{x) by £c — «! is a polynomial of degree n —1. Suppose that

aoic" -f- «ia;«- 1 -f ••• + «« = «o(^ - «^i) (ic"" ^ + ^i^c""^ H h ^„- 1).

By our assumption the quotient a;"" ^ -f Jicc'*"^ 4- • • • + ^„_ i =
has at least one robt, say az, to which corresponds the factor
X — a^. Thus

f{x) =aQ{x- a^ {x — a^) (a;"-^ -\- CiX"-^ -\ h c^.g).

Proceeding in this way we find successive roots and corre-
sponding linear factors until the polynomial is expressed as the
product of n linear factors as follows :

f{x) =ao(x- ai) (x-a2)'"(x- a„) = 0,
where the roots are Uu ag, • • • , a„.

Remark. This theorem gives no information regarding how many of the roots
may be real or imaginary. This depends on the particular values of the coefl&cients.

THEORY OF EQUATIONS 173

Corollary. Any polynomial in x of degree n may he expressed
as the product of n linear factors of the form x — a, where a is
a real or a complex number.

It should be noted that the roots are not necessarily distinct.
Several of the roots and hence several of the factors may be
identical.

If f(x) is divisible by {x — aiY, that is, if a^ = a^, we say that
oTi is a double root of the equation. Similarly, \i f(x) is divis-
ible by {x — oTi)'', a^ is called a multiple root of order r. When
we say an equation has n roots we include each multiple root
counted a number of times equal to its order.

Theorem. An equation of degree n has no more than n
distinct roots.

Let/(x) = ^0^" H f" ^n = ^ have the roots a-^^, cc2j'"j *».• Write

the equation in the form

ao(x — ai) • • ' (x — a^) = 0.

If r is a root distinct from a^, •••,«:„, it must satisfy the equation
and

ao(r-aj)"'(r-a^) = 0.

Since this numerical expression vanishes one of its factors
must vanish (§5). But r =f= ai, thus r — a^ =^ 0. Similarly, no
one of the binomial factors vanishes. Thus (§5) «<, — 0, which
contradicts the hypothesis that the equation is of degree n.

This theorem may also be stated as follows :

Corollary I. If an equation a^x" + a^^x"" "^ + \- a^= of

degree n is satisfied by more than n values of x, all its coefficients
vanish.

The proof of the theorem shows that if the equation has n + 1
roots, ao = 0. We should then have remaining an equation of
degree n —1, also satisfied hj n +1 values of x. Thus the coeffi-
cient of its highest power in x vanishes. Similarly, each of the
coefficients vanishes.

174 ' ADVANCED ALGEBRA

Corollary II. If two ^polynomials in one variable are equal
to each other for every value of the variable^ the coefficients of
like ^powers of the' variable are equal and conversely.

Let a^x"" + a^x""-^ H h o^„ = ^o^" + ^i^c""^ H h *„

for every value of x.

Transpose, {a^ — h^x''-\--'--\-a^ — h^ = ^.

By Corollary I, a^ — h^^^ 0, or a^ = b^,

ai — bi = 0, or a^ = bi,

(^n-K = 0, or a„ = ^>„.

167. Graphical interpretation. The graphical interpretation
of the theorems of the preceding section is that the graph of an
equation of degree n cannot cross the X axis more than n times.
Since each crossing of the X axis corresponds to a real root, there
will be less than n crossings if the equation has imaginary roots.

168. Imaginary roots. We now show that imaginary roots
occur in pairs. This we prove in the following

Theorem. If a -\- ib is a root of an equation with real coeffi-
cients, a — ib is also a root of the equation.

li a -{- ib is a root of the equation <Xo^" + a^x^' ^ -\- -••-{- a^ = 0^
then X —(a -{- ib) is a factor (p. 166). We wish to prove that
X — {a — ib) is also a factor, or what amounts to the same thing,
that their product

[x —{a-\- ib)'\ [x —{a — lb)'] = [(x — a) — ib"] [(x — a)-\- ib"]

is a factor of f(x). Divide f(x) by (x — a)^ + b^ and we get

f(x) = Q(x) [(x - af J^b^-]J^rx + r', (1)

where r and r' are real numbers. This remainder rx + r' can be
of no higher degree in x than the first, since the divisor

(x - af + b''

THEORY OF EQUATIONS

175

is only of the second degree (§ 26). Now this equation (1) being
an identity is true whatever value is substituted for x, as, for
instance, the root of /(x), a -\- ib. Substituting this value for x,
we get i

f{a Jrib) = = Q{a + ib) \_{a + ib - ay + b^-]^^ r {a -\- ib) + r',

or (p. 33) and (p. 3) = + r«^ -f r' + irb,

or (p. 156) m + r' = 0, (2)

rb = 0. (3)

Since b ^ 0, by (3), p. 3, r = 0.

Also from (2), r' = 0. >

Hence rx -{- r' = 0.

Consequently there is no remainder to the division of f(x) by
(x — a)^ -\- U\ and hence if a + ib is a root oif(x), a — ib is also a
root.

Corollary. Every equation of odd degree with real coeffi-
cients has at least one real root.

The roots cannot all be imaginary, else the degree of the equa-
tion would be even by the preceding theorem.

169. Graphical interpretation of imaginary roots. When we
plot the equations

y = x^ + Ax''-4. (1),
Yk

y = x^-i-4:x^-l (2),

176

7/ = ^^ + 4^2 (3),

y

=

a;8

+ 4

x"

+1 (4

0.

^

\

/

/

\

/

/

>v

/

r

\

/

^

\

\

\

\

\

\

J

x^

,

—

—

X

we see that corresponding to the increase of the constant term is
a corresponding elevation" of the curve with respect to the X axis.
In fact in each case the curve is the same, but the value of y is
gradually increased. In (1) and (2) we have three real roots, in
(3) the curve touches the X axis, and in (4) we have only one
real root. As the elbow of the curve is raised and fails to intersect
the X axis a pair of roots cease to be real, and since a cubic equa-
tion always has three roots, a pair of roots become imaginary.
Thus we have the

Pkinciple. Corresponding to every elbow of the curve that
does not intersect the X axis there is a pair of imaginary roots
of the equation.

The converse is not always true. It is not always possible to
find as many elbows of the curve which do not meet the X axis
as there are pairs of imaginary roots.

EXERCISES

Plot the following equations and determine from the plot how many roots
are real.

1. X* - 1 = y. 2. a;6 - 2 = y. 3. x^ - x - 1 = y.

4. x4 + l = y. 5. x* + x + l = ?/. 6. x* + 2x2 + 2 = y.

7. x8 - 3x2 - X + 1 = y. 8. x8 - 2x2 + 4x - 1 = y.

9. 2x8 + 3x2 + 6x + 6 = y. 10. x^ - 3x2 - 4x - 5 = y.

THEORY OF EQUATIONS 177

170. Relation between roots and coefficients. If we write the
expression (Corollary, p. 173)

and multiply the factors, we obtain by equating coefficients of like
powers of x (p. 174) relations between the roots and the coeffi-
cients. Take for example n = S.

x' + Wx^ + hx + b, = (x- 13,) (x - 13,) (X - p,)

= x'- (13, -{-/3, + 13,) x' + ()8i/82 + ^2^3 + ft A) X - )8i A^3 = 0.
. Hence j^ = _ (/3^ + ^^ + ^3),

*2 = Aft + Aft + A A,

^3 = -AAft-

This suggests the

Theorem. The coefficient of ^""^ is equal to the sum of the
roots with their signs changed.

The constant term is equal to the product of the roots with
their signs changed.

In general the coefficient of a?""*" is equal to the sum of all
possible products of r of the roots with their signs changed.

We prove this theorem by complete induction.
First. We have already established the theorem for equations
of degree two on p. 106 and for equations of degree three above.
Second. Assume the theorem for n =^m. That is, if

£c- + ^'1^'"-' + • • • + *«. = (^ - A) (^ - A) • • • (^ - A). (1)
we assume that h^, the coefficient of ic'"-'", is the sum of all possi-
ble products of r of the numbers — A? — A? ' • '? ~ A-

Multiply both sides of (1) hy x — A« + i. Denote the result by
^m+i + b^^x^ + . . . + 5'^^^ = (x- p,)(x ^p,)...(x- ^^ + 0- (2)

The term in 0;"'+^-'" in this equation is obtained by multiplying
the terms b.x'^-'- and &^_ia;'»+^-'' in (1) by x and - p^+i respec-
tively. That is, in (2)

b'r = K + b^-,(-(im^O' (3)

Now all possible products of r of the quantities — A> —ft*
.. •, — A 4.1 may be formed as follows : (1) Neglect — ft + i, and
form all possible products of r of those remaining. The sum of

these is b^. (2) Form all possible products of r - 1 of — )8i
— Aj • • •> — Pm^ not including — Pm + u and multiply each product
^y — Pm+\- Add all the products obtained. This process, it is
observed, is precisely that indicated by (3).

Remark. It is noticed that in the rule the signs of the roots are always changed
before forming any term. This does not involve any change when r is an even
number, but is included in the rule for the sake of uniformity.

Corollary. Every root of an equation is a factor of its con-
stant term.

171. The general term in the binomial expansion. On p. 129

we gave an expression for the (r + l)st term of the binomial ex-
pansion, the validity of which we now establish. In (1), § 170,
let ySi = ^2 = • • • = Pn- Denote this common value by — a. The
expression (1) becomes, on writing n in place of m,

X- + hx""-' + --- + b,=(x + ay.
By the theorem in § 170, b^ is the sum of all possible products
of r of the negative roots. Since there are

a.=

n(n — 1) "• (n — r -i-l)

ri

such products, and since the roots are now identical, we obtain
n(n-l)>-'(n-r-{-l) ^^_^^^
rl
as the form of the (r + l)st term of the expansion of (x + a)".

172. Solution by trial. Since by the previous corollary every
root of an equation is a factor of its constant term, we may in
many cases test by synthetic division whether or not a given equa-
tion has integral roots. Thus the integral roots of the equation

ic* _ 8a;» + 4:ic=» -f 24a; - 21 = (1)

must be factors of 21.

We try + 1 by synthetic division, '

1_8h-4 + 24- 2111

-|.1_7_ 3-1-21
1_7_3-|-21
Thus 1 is a root of (1), and the quotient of the equation by
"^-^^^ a;»-7x='-3a;-f 21 = 0. (2)

THEORY OF EQUATIONS 179

If this equation has any integral root it must be a factor of 21.
We try + 3 by synthetic division,

1_7_ 3 + 21|3

+ 3 _ 12 - 45
1-4-15-24

Thus 3 is not a root. We try + 7,

1_7_3 + 21|7
+ 7 4- - 21

1+0-3
Thus 7 is a root, and the remaining roots of (1) are the roots of
a;2 - 3 = 0,
that is, X =± V3.

Hence the roots of (1) are + 1, + 7, ± V3.

EXERCISES

Solve by trial :

1. x3 - 7x2 + 50 = 0. 2, x3 - 9x 4- 28 = 0.

3. x3-36x- 91 = 0. 4. x3 + 9x + 26=:0.

5. x8 - 19x + 30 = 0. 6. x3 - 27x - 54 = 0.

7. x8 + 2x2 - 23x + 6 = 0. 8. x3 - 6x2 + 11 X - 6 = 0.

9. x3 - 2x2 - llx + 12 = 0. 10. x3 - 8x2 ^ 19a; _ 20 = 0.

11. x8 + 9x2 + 27x + 26 = 0. 12. x* - 8x3 + 8x2 ^ 40x - 32 = 0.

13. X* - 13x2 + 48x - 60 = 0. 14. x* - 3x3 - 34x2 + 18x +168 = 0.

15. X* +8x3- 7x2 -50x + 48 = 0.

16. x* - 3x3 - 5x2 + 29x - 30 = 0.

17. x* - 6x3 + 13x2 - 30x + 40 = 0.

18. x4-8x3+21x2-34x + 20 = 0.

19. x* - 12x3 + 43x2 - 42x + 10 = 0.

173. Properties of binomial surds. A binomial surd is a

number of the form a ± V^, where a and b are rational numbers,
and where b is positive but not a perfect square.

Though we have not explicitly defined what we mean by the sum of an
irrational number and a rational number, we shall assume that we can
operate with the binomial surd just as we would be able to operate if b were
a perfect square.

Theorem \. If a binomial surd a + V^ = 0^ then a =0 anc
b = 0.

J,i a-\- V^ = and either a = ov b = 0, clearly both must equal
zero. Suppose, however, that neither a nor b equals zero. Then
transposing we have a =^ 'Vb, and a rational number would be
equal to an irrational number, which cannot be. Hence the only
alternative is that both a and b equal zero.

Theorem II. If two binomial surds, asa -\- -sib and c + V5, are
equal, then a = c and b = d.

Let a + V^ = c -f V^.

Transposing c, a — c -\- V^ = V5. (1)

Square and we obtain

(^ci - cy-\-b -\- 2(a - c)Vb = dy
or

(^a, - cy + b - d -\- 2(a - c) Vb = 0.

Thus, by Theorem I, either b = 0, which is contrary to the defi-
nition of a binomial surd, or a ~ e = 0, that is, a = c. In the
latter case (1) reduces to Vb == Vd, OTb = d, and we have a = c and
b = dy which was to be proved.

a 4- Vb and a — V^ are called conjugate binomial surds.

Theorem III. If a given binomial surd a + Vb is the root of
an equation with rational coefficients, then its conjugate is also a
root of the same equation.

The proof of this theorem, which should be performed in writ-
ing by each student, may be made analogously to the proof of
the theorem on p. 174.

174. Formation of equations. If we know all the roots of an
equation, we may form the equation in either one of two ways
(see p. 167 and p. 177).

First method. If a^, oc^,-", oc^ are the given roots, multiply
together the factors x — a^,- ■•,x — a^.

Second method. From the given roots form the coefficients
by the rule on p. 177.

THEORY OF EQUATIONS 181

If the equation and all but one of its roots are known, that
root can be found by the solution of a linear equation obtained
from the coef&cient of the second or the last term. If all but two
of its roots are known, the unknown roots may be found by the
solution of a pair of simultaneous equations formed from the
same coefficients.

In the solution of the following exercises use is made of the
theorem on p. 174, Theorem III, p. 180, and the various relations
between the roots and the coefficients.

EXERCISES

1. Form the equations which have the following roots. Check the process
by using both methods of § 174.

(a) 2, - 3, 1.
Solution :

First metfiod. {x - 2){x + 3) (x - 1) = JC^ - 7 x + 6.

Second method. Let the equation be

x3 + 6ix2 + biX + 63 = 0.
Then, by § 170, 61 = - (2 - 3 + 1) = 0, - • \

62 = -6 + 2-3 = -7,

63 = -2-3. -1 = 6.
The equation then is x^ — 7x + 6 = 0.

(b) 1, 2, 3. (c) 2, 2, 2, 2.
(d) 3, 1, 1, 0. (e) 1, 0, 0, 0.
(f) ±V2,±i. (g) 2,4, -6.
(h) 2, - 3, 1, 0. (i) 2, 3, - 6.

. (j) 7, V5, -V5. (k) 1,2, -1, -|.

(1) 3, 1 + i, 1 - i (m) _ 4, - 3, 3 ± V5.

(n) 1 ± i, - 1 ± i (o) 2, V=^, -V^^.

(p) -1,2,3,-4. (q)2^,3|, -H, -2^.
(r) ±V6, ±iV7. (s) -5, 2 + V6, 2-V5.

(V) 3, ^i±^, i:!^. (w) - 1, l±i^^ 1^.

182

2. The equation x* + 2x3 - 7^2 _ 8x + 12 = has two roots — 3 and
+ 1. Find the remaining roots.

Solution : Let the unknown roots be a and h.
Then, by § 170, _a-6 + 3-l = 2,

- 3 a6 = 12.
Solving for a and 5, we obtain a = — 2 or +2,

6 = + 2 or - 2.

3. x8 — 7x + 6 = has the roots 2 and 1. Find the remaining root.

4. X* — 3x + 2 = has the root 1. Find the remaining roots.

5. x^ — 18 X — 35 = has the root 5. Find the remaining roots.

6. Two roots of x* — 35 x^ -f- 90 x — 56 = are 1 and 2. Find the remain-
ing roots.

7. The roots of x^ - 6 x2 - 4 x + 24 = are in A. P. Find them,

8. The two equations x3-6x2 + llx-6 = and x^ -14x2 + 63 x - 90 =
have a root common. Plot both equations on the same axes, and find all the
roots of both equations.

9. Determine the middle term of the equation whose roots are — 2,
+ 1, 3, — 4 without determining any other term.

10. What is the last term of the equation whose roots are — 4, 4, ± V— 3 ?

11. One root of x* - 4x8 + 5x2 + 2x + 52 = is 3 - 2i. Find the remain-
ing roots.

12. One root of x* - 4 x^ + 5 x2 + 8 x - 14= is 2 + i V3. Find the others.

13. Plot the following equations, determine all the integral roots, and
find the remaining roots by solving.

(a) X* - 6x8 + 24x - 16 = 0.

X

y

-1

- 33

-2

-3

+ 155

In this plot two squares on the X axis represent a unit of x, while one
square on the Y axis represents ten units of y. The integral factors are
X — 2 and x + 2, since ± 2 are roots, that is, are values of x for which the

THEORY OF EQUATIONS 183

curve is on the X axis. To find the quotient of our equation we first divide
synthetically by 2, and then the quotient by — 2, using the principle given
in § 161.

1-6+ + 24- 16 12
+ 2 - 8-16 + 16
^ 1-4- 8+ 8 |-2

- 2 + 12 - 8
1-6+ 4

Thus the quotient of the polynomial and (x — 2) (« + 2) is «2 __ ga; 4. 4
Solving the equation

x2-6x + 4 = 0,

we obtain the two remaining roots, x = 3 ± V5. These remaining roots
might also be found by the method of exercise 2.

(b) x3 - 6x - 12 = 0. (c) x3 - 8x2 + 7 = 0.

(d) x8 - 7x2 + 50 = 0. (e) x^ - 8x2 + 13x- 6 = 0.

(f) x8-6x2+7x-2 = 0. (g) x3 + 3x2 + 4x- 24 = 0.

(h) X* - 3x3 + 7 x2 - 21 X = 0. (i) X* - 3x3 - 7 x2 + 27 x - 18 = 0.

'(j) x*-9x3 + 21x2-19x + 6 = 0.

(k) How many imaginary roots can an equation of the 5th degree have ?
(1) x3 — ax2 + 6x + c = has two roots whose sum is zero. What is the
third root ? What are the two roots whose sum is zero ?

(m) x3 + x2 + 6x + c = has one root the reciprocal of the other. What
are the values of the roots ?

(n) x3 — 4x2 + ox + 62 = has the sum of two roots equal to zero. What
must be the values of a and h ?

(o) X* - 3 x3 + 6x + 9 = has the sum of three of its roots equal to zero.
What must be the value of 6 ?

175. To multiply the roots by a constant. Suppose we have
the equation

fix) = a,x- +- a^x--' + . . . + «^ = 0, (1)

whose roots are «ri, a^, • • • , «:„. An equation of this type for
values of n greater than 2 is usually not solvable by elementary
methods. It often happens, however, that by changing its form
slightly we may obtain an equation one or more of whose roots
we can find. We shall see that if an equation has rational roots
we may always find them if we change the form of the equation
as indicated on the following page.
/

We seek to form from (1) an equation whose roots are equal
to the roots of (1) multiplied by a constant factor, as k. Thus
the equation we seek must have the roots kai, ka^, ka^. We
carry out the proof, which is perfectly general, on the equation
of the third order

f(x) = a^x^ -f- a-^x^ -f «2^ + ^3 = 0,

whose roots are a^, a^, a^. The equation that we seek must have
roots kai, ka^, ka^. Since now (§ 53) f{x) = is satisfied by
a, where a stands for any one of the. roots, that is, since f(a) = 0,

evidently /(t) = is satisfied by ka, that is.

Hence we obtain an equation that is satisfied by ka^, ka^^ ka^,
if in/(£c) we let a; = t*

The required equation is then

Akj-'W^l^^-k^''^-^^

or, multiplying by k^,

a^z^ + ka-^z'^ -{- k^azZ -{- k^a^ — 0.
This affords the general

KuLE. To multiply the roots of an equation hy a constant k,
multiply the successive coefficients beginning with the coefficient of
x""'^ hy k,l^, • • -,1^ respectively.

In performing this operation the lacking powers of x should be
supplied with zero coefficients.

Example. Multiply the roots of 2 x^ — 3 x + 4 = by 2.
Multiply the coefficients by the rule above,

2«8 + 2-0x2-4-3x + 8-4 = 0.
Simplifying, »» - 6 x + 16 = 0.

THEORY OF EQUATIONS 185

When an equation in form (2), p. 166, has fractional coefficients,
an equation may be formed whose roots are a properly chosen
multiple of the roots of the original equation and whose coeffi-
cients are integers.

Corollary I. When k is a fraction this method serves to
divide the roots of an equation hy a given number.

Corollary IL When k = — 1 this method serves to form an
equation whose roots are equal to the roots of the original equa-
tion hut opposite in sign. This is equivalent to the statement
that /(— x) = has roots equal hut opposite in sign to those

• EXERCISES

1. Form the equation whose roots are three times the roots of

Solution : Supplying the missing term in the equation, we have

«* -6x3 + 0x2- X + 1 = 0.
Since A; = 3, we liave by the rule

x* -3 -6x3 + 9 -0x2- 27 -x + 81 = 0, 'r

or x*-18x3 -27x + 81 = 0. '

2. Find the equation whose roots are twice the roots of

x* + 3x3-2x + 4 = 0,

3. Find the equation whose roots are one half the roots of

x3-2x2 + 3x-4 = 0.

4. Find the equation whose roots are two thirds the roots of

x3-4x-6 =0.

5. By what may the roots of the following equations be multiplied so thai
in the resulting equation the coefficient of the highest power of x is unity
and the remaining coefficients are integers ? Form the equations.

(a) 3x3- 6x + 2 = 0.

Solution : We wish to bring into every term such a factor that all the
resulting coefficients are divisible by 3.

Let A: = 3.

Supply the lacking term,

3x3 + 0x2 -6x + 2 = 0.

By rule, 3x3 + 3 • 0x2 - 9 • 6x + 27 • 2 = 0.

• Dividing by 3, x3 - 18 x + 18 = 0.

186

x2

(b) X3 + - -

1 = 0.

(C) X8 - i = 0.

(d) «» + 7x2 + -x + - = 0,

62

6»

(e) x4 + ^

x2

+ 1 = 0.

(f) 2x8 - 3x2 -x + 4 = 0.
(h) x*-6x3-2x2+ 1 =0.

(g) 3x4-3x2-4x + l = 0.
(i) 16x4 - 24x3 + 8x2 -2x + 1 = 0.

6. Form equations whose roots are the negatives of the roots of the fol-
lowing equations.

(a) x8-4x + 6 = 0.

Solution : Supply the lacking term,

x3+ 0x2-4x + 6 = 0.
Changing signs we obtain by Corollary II

x3_0x2-4x- 6 = 0,
or x3 - 4 X - 6 = 0.

(b) X' - 2x2 - 4x = 0. (c) x* - 3 x2 + 1 = 0.

(d) X* - 2x3 + x2 4- 2x - 1 = 0. (e) x3 + 3x2 + 7x - 13 = 0.

7. What effect does changing the sign of every term of the member
involving x have on the graph of an equation?

8. What is the graphical interpretation of the transformation which
changes the signs of the roots of an equation, that is, what relation does
the graph of the equation before transformation bear to the graph of the
equation after transformation (a) when the degree is an even number,
(b) when the degree is an odd number?

9. If 4x4 - 16x3 - 86x2 + 4x + 21 = has as two roots - ^ and - 3,
what are the roots of 4x4 + 16x3 - 85x2 - 4x + 21 = 0?

10. If a root of xs - 11 x2 + 36 x - 36 = is 2, what are the roots of
x»+ 11x2+ 36x + 36 = 0?

176. Descartes' rule of signs. A pair of successive like signs in
an equation is called a continuation of sign. A pair of successive
unlike signs is called a change of sign.

In the equation

2x4_3a;8 + 2x2 + 2a;-3 =

(1)

are one continuation of sign and three changes of sign. This may be seen more
clearly by writing merely the signs, + — + + —.

Let US now inquire what effect if any is noted on the number
of changes of sign in an equation if the equation is multiplied by

THEORY OF EQUATIONS 187

a factor of the form x — a when a is positive, that is, when the
number of positive roots of the equation is increased by one.
Let us multiply equation (1) by a; — 2. We have then

x-2

2x^ -3x*-{-2x^-{-2x''-Sx

-4:X^+6x^-4:X^-4:X-\-6

2x' - 7 x^ -{- Sx^ - 2x^ - 7 x -{- 6

In this expression the succession of signs is + — + — — +,
in which there are four changes of sign, that is, one more change
of sign than in (1). If an increase in the number of positive roots
always brings about at least an equal increase in the number of
changes of sign, there must be at least as many changes of sign in
an equation as there are positive roots. This is the fact, as we
now prove.

Descartes' rule of signs. An equation f(x) = has no more
real positive roots than f(x) has changes of sign.

Illustration. In the equation of degree one £c — 2 = there
is one change of sign and one jjositive root. In the case of a linear
equation there is no possibility of more than one change of sign.
In the quadratic equation x'^-\-2x-\-l = Q there is no change of
sign, and also no positive root since for positive values of x the
expression x^ -f 2 cc + 1 is always positive and hence never zero.
In the equation ic^ + 2ic — 3 = we have one change of sign,
and one positive root, + 1.

We shall prove this general rule by complete induction.

First. W^e have just seen that the rule holds for an equation
of degree one.

Second. We assume that the rule holds for an equation of
degree m, and prove that its validity for an equation of degree
m -\-l follows. We shall show that if we multiply an equation
of degree m hj x — a, where a is positive, thus forming an equa-
tion of degree m + 1, the number of changes of sign in the new
equation always exceeds the number of changes of sign in the

188

original equation by at least one. That is, the number of chanj
of sign increases at least as rapidly as the increase in the number
of positive roots when such a multiplication is made.

Let/(£c) = represent any particular equation of the nth. degree.
The first sign of f(x) is always +. The remaining signs occur in
successive groups of + or — signs which may contain only one
sign each. If any term is lacking, its sign is taken to be the same
as an adjacent sign. Thus the most general way in which the
signs of /(x) may occur is represented in the following table,
in which the dots represent an indefinite number of signs. The
multiplication of f(x) by a; — a is represented schematically, onl^P"-^
the signs being given. ^

fix)

X — a

All + signs
+ •••• +

All - signs

All + signs

All - signs

Further
groups

+ •••• +

All - signs

+ -

xf{x)
- ocfix)

+ + ••• +

- + ••• +

+ + ••• +
+

- + ••• +

+ +••• +
+

-+ + +

{x-a)f{x)

+ d=---±

-±.-.±

+ ±---±

-±...±

+ ± ±

-±•••±4-

The ± sign indicates that either the + or the — sign may occur
according to the value of the coefficients and of a. The verti-
cal lines denote where changes of sign occur in f(x). Assuming
that all the ambiguous signs are taken so as to afford the least
possible number of changes of sign, even then in (x — cc)f(x)
there is a change of sign at each or between each pair of the
vertical lines, and in addition, one to the right of all the vertical
lines. Thus as we increase the number of positive roots by one
the number of changes of sign increases at least by one, perhaps
by more.

The only possible variation that could occur in the succession
of groups of signs in f(x), namely, when the last group is a
group of 4- signs, does not alter the validity of the theorem.

THEORY OF EQUATIONS 189

We illustrate the foregoing proof by the following particular
example.

Let f{x) = x^ -4:X^ - x-{-2, and let a = 2.

Multiply f(x), 1 + 0-4 + 0-1 + 2 4 changes

by x-2, 1-2

xf(x), 1 + 0-4 + 0-1 + 2

-2f(x), -2-0 + 8-0 + 2-4

(x~2)f(x), 1-2-4 + 8-1 + 4-4 5 changes

177. Negative roots. Since /(—a:) has roots opposite in sign
to those of f(x) (p. 185), we can state

Descartes' rule of signs for negative roots. f(x) has nc
more negative roots than there are changes in sign in /(— x).

If by Descartes' rule it appears that there cannot be more than
a positive roots and b negative roots, and if a -{- b <n, the degree
of the equation, then there must be imaginary roots, at least
n — (a -{- b) in number.

EXERCISES

1. Prove Descartes' rule of signs for x^ + bx + c = directly from the
expression for b and c in terms of the roots (see § 115).

2. Find the maximum number of positive and negative roots and any
possible information about imaginary roots in the following equations.

(a) x3 + 2 x2 + 1 = 0.

Solution : Writing signs of f{x), + + +, there is no change, hence no
positive root.

Writing signs of /(— £c), — + + , there is one change, hence no more
than one negative root. Since there can be only one real root there must
be two imaginary roots.

(b) x3 + 1 = 0. (c) X* - 2 = 0.

(d) x3 - X + 1 = 0. (e) x6 - X + 1 = 0.

(f ) x4 + X + 1 = 0. (g) x6 + x2 + 1 = 0.

(h) x3-6x2+4x-l = 0. (i) x5-2x4-3x3+4x2+x + l = 0.

(i) x6 + 2x* - 6x3 - 4x2 + X - 1 = 0.

178. Integral roots. In finding the rational roots of an equa-
tion we make use of the following

Theorem. If the equation

^ + «i^"-'+---+a„ = ^ (1)

{where the oUs are integers) has any rational root, such root must
he an integer.

Suppose - be a fraction reduced to its lowest terms which
satisfies the equation.

Then ^ + 2l£r! + ... + „„ =

is an identity.

Then clearing of fractions and transposing,

f = --q {ciip"~^ H h a„!?"~0-

Thus some factor of g' is a factor of ^^ that is, oip (p. 52), which
contradicts the hypothesis that - is reduced to its lowest terms.

Thus all the rational roots of the equation are integers, which
as we know (§ 170) are factors of a„.

179. Rational roots. If we seek the rational roots of

aox"" H [- a„ = 0,

where a^ =^ 1, we can multiply the roots by a properly chosen
constant (§ 175) and obtain an equation of form (1) above whose
integral roots may easily be found by synthetic division.

Example. What rational roots, if any, has

8x»+lla;-14 = 0? (1)

Multiply the roots by 8, 3 a^ + 99 x - 378 = 0.

Divide by 3, x^ + 33 x - 126 = 0. (2)

Since by Descartes' rule of signs equation (1) has only one positive root
and no negative root, we do not need to carry the table further than to test
for a positive root.

THEOKY OF EQUATIONS 191

y Form a table of values for equation (2) by synthetic division.

~^ We need only to try the factors of 126 (§ 170).
_ g2 Thus (2) has the root 3. Hence the original equation (1) has

Q the root 3^3 = 1.

KuLE. To find all the rational roots of an equation, trans-
form the equation so that the first coefficient is + 1.

Find the maxiymum number of positive and negative roots by
Descartes' rule of signs.

Find the integral roots of this equation by trial, and the
roots of the. original equation by dividing the integral roots
found by the constant by which the roots were multiplied.

By the Theorem § 178 we are assured that all the rational
roots can be found in this way.

EXERCISES

Find all the rational roots of the following equations.

1. 4x8 = 27(x + l). 2. 15jc8 + 13a;2-2 = 0.

3. 4a;3-6x-6 = 0. 4. x^- 2f x2 + 2|x - 1 = 0.

5. 4x3 - 8x2 - X + 2 = 0. 6. 3x* - 8x8 _ ^q^2 + 25 = 0.

7. 4x8-4x2+ x-6 = 0. 8. 3x8+ 13x2+ llx - 14 = 0.

9. 4x8 + 16x2 -9x- 36 = 0. 10. 2x3 - 21x2 + 74x - 85 = 0.

11. 6x8 - 47x2 + 71X + 70 = 0. 12. 12x8 - 52x2+ 23x + 42 = 0.

13. 6x8-29x2+ 53x- 45 = 0. 14. 6x*- x8- 8x2- 14x + 12 = 0.
15. 27x3+ 63x2+ 30x- 8 = 0. 16. 2x*-13x8 + 16x2- 9x +20 = 0.
17. 3x8 - 26x2 +52x- 24 = 0. 18. Ox* - x3 - 49x2+ 55x - 50 = 0.
19. 18x8+ 81x2+ 121x + 60 = 0. 20. 12x4 + 6x8 - 24x2- 9x + 9 = 0.

21. 10x4 + 18x8 - 16x2 + 8x- 20 = 0.

22. 9x4 + 15x8 - 143x2 +41X + 30 = 0.

23. 36x4 - 72x8 - 31 x2 + 67 x + 30 = 0.

24. 24x4- 108x8+ 324x2 -240X + 60 = 0.

180. Diminishing the roots of an equation. In the preceding
sections we have solved completely the problem of finding the
rational roots of an equation. We now pass to the problem of

V

finding the approximate values of the irrational roots of an eqnar
tion. In carrying out the process that we shall develop it is
desirable to form an equation whose roots are equal respectively
to the roots of the original equation each diminished by a constant.

Let fix) = aox"" + a^x''-'^ -\ h «« = 0, (1)

whose roots are ^i, a^, • ••, a^. Let a be any constant. We seek
an equation whose roots are a^ — a, a^ — a, - • • , a^ — a. ^~^^^v_

If we let a stand for any one of the roots of (1), since f(a) =
(p. 33), we see that

/(« + a) = is satisfied by a — a,
that is, f{a-a-\-a)= f(a) = 0.

Thus to form the desired equation replace x by z -\- a. We
obtain

f(x) = f(z + a) = ao(z + a)» ■}- a,{z -{- af-^ + . . . + a„ = 0.

Developing each term by the binomial theorem and collecting
like powers of z, we get an equation of the form

fix) = Fiz) = A,z- + A^z-^ + . . . + ^^ = 0, (2)

where the ^'s involve the a's and d. This is the equation desired.
We now seek a convenient method of finding the values of the
coefficients ^o? ^i? ^2> •••> ^n when a^^ %, a^^ •••, a„ are given
numerically. Now A^ is the remainder from the division of Fiz)
by z. But since Fiz) =fix) and z = x — a, the remainder from
dividing Fiz) by z is identical with' the remainder from dividing
fix) hj x — a. Thus A^ is the remainder from dividing fix) by
X — a. Furthermore, since A^_y^ is the remainder from dividing

—^^ by z, it is also the remainder from dividing ^^-^^

by a; — a. The process may be continued for finding the other ^'s.
We may then diminish the roots of an equation by a as follows :

EuLE. The constant term of the new equation is the remaivr-
der from dividing fix) hy x — a.

i

THEORY OF EQUATIONS

193

The coefficient of z in the new equation is the remainder from
dividing the quotient just obtained hy x — a.

The coefficients of the higher powers of z are the remainders
from dividing the successive quotients obtained hy x — a.

Example. Form the equation whose roots are 2 less than the roots of
x4 _ 2 x3 - 4 x2 4- x - 1 = 0.

The divisions required by the rule we carry out synthetically (p. 169).
1-2-4+1- 1[2
+ 2 +0 -8 -14
-15

1 +

+ 2

-4 -7

+ 4 +0

1 + 2 +0
+ 2 +8

-7

1 + 4

+ 2

+ 8

1+6

The desired equation is

^4 + 6x3 + 8x2

7x-15 = 0.

181. Graphical interpretation of decreasing roots. If an equa-
tion has roots a units less than those of another equation, if a is
positive its intersections with the X axis or with any line parallel
to the X axis are a units to the left of the corresponding inter-
sections of the first equation. It is, in fact, the same curve, except-
ing that the Y axis is moved a units to the right. If a is negative,
the Y axis is moved to the left.

EXERCISES

Plot, decrease the roots by a units, and plot the new axes.

1. x4-3x3 -2x-3 = 0. a = 3.

Solution: 1-3- 0- 2-3[3

(1)

+ 3-0

-

-6

1_0 - - 2
+ 3 + 9+27

-9

1 + 3 + 9

+ 3 + 18

+ 25

1 + 6

+ 3

1 + 9

+ 27

194

Thus the required equation is

0. (2)

X

y

- 3

: 1

- 7

2

-16

3

- 9

-1

+ 3

3. x3 _ 8 = 0. a = 1.4.
5. x3 + 4x-8 = 0. a = 3.
7. x3 + 2 X + 6 = 0. a = -\.
9. «8-2x2 + 8x-7 = 0. a = 2.

11. x4-3x2 + 2x

12. X

13. 2x3-6x2 + 4x -3 =

14. X* + 6x3 + 10x2 + x - 1 =

In the figure one square on the Faxis repre-
sents two units of y, and two squares on the
X axis represent one unit of x.

2. x*-16 = 0. a = 2.
4. x4- 2x2 + 1 = 0. a = .2.
6. x3-4x2-2 = 0. a = .5.
8. x3 + 4 x2 + X - 6 = 0. a = - .4.
10. x3-3x2 + x-l = 0. a = -.3.
2 = 0. a = -2.
15x2 + 7x + 125 = 0. a = 5. -

a = -3.

a = -l.

182. Location principle. If when plotting an equation y =f(x)
the value x = a gives the corresponding value of y positive and
equal to c, while the value x = b gives the corresponding value
of y negative, say equal to — d, then the
point on the curve x = a, y = c is above
the X axis, and the point on the curve
x = byy = — -^ is below the X axis. If our
curve is unbroken, '^ must then cross the
X axis at least once between the values
X — a and x = h^ and hence the equation
must have a root between those values
of x. The shorter we can determine this
interval a to b the more accurately we can find the root of the
equation. This property of unbrokenness or continuity of the
graph of ?/ = ^0^" 4- aix"~^ + • • • + a„ we assume. We assume
then the following

THEORY OF EQUATIONS

195

Location principle. When for two real unequal values of x,
x = a and x = h, the value of y —f{x) has opposite signs,
the equation f{x) = has a real root between a and h.

Illustration. The equation f(x) =ic^-f3cc — 5 = has a
root between 1 and 2. Since /(I) = - 1, /(2) = 9.

183. Approximate calculation of roots by Horner's method.

We are now in a position to compute to any required degree of

accuracy the real roots of an equation. Consider for example

the equation ^s + 3,_20 = 0. (1)

Form the table of values for plotting the equation

ic8 + 3 a: - 20 = ?/.

By the location theorem we find
that a root is between + 2 and + 3.
To find more precisely the position
of the root we might estimate from
the graph the position of the root
and substitute say 2.3, 2.4, and so 3+16

on, until we found two values be-
tween which the root lies. We can
gain the same result with much _ i _ 24
less computation if we first dimin-
ish the roots of the equation so that
the origin is at the less of the two integral values between which
we know the root lies. Here we decrease the roots of (1) by 2,

1 4- + 3 - 20[2

y
-20

-16

+ 2 + 4+14

1 + 2 + 7
+ 2 -h 8

- 6

1 + 4
+ 2

+ 15

1+6

The equation whose roots are decreased by 2 is
a:» + 6ic2 + 15a;^6 = 0.

(2)

We know that (2) has a root between and 1, since equation (1)
has a root between 2 and 3. From the graph we can estimate the
position of the root. Having made an estimate, say .3, it is neces-
sary to verify the estimate and determine by synthetic divi^io^
precisely between which tenths the root lies. Thus, trying .3, we"

^^*^^^ 1 + 6.0 + 15.00 - 6.000[_^

+ 0.3+ 1.89 + 5.067
1 + 6.3 + 16.89 - 0.933 ()

which shows that for x = .3 the curve is below the X axis, hence
the root is greater than .3. But we are not justified in assuming
that the root is between .3 and .4 until we have substituted .4
for X. This we proceed to do.

1 + 6.0 + 15.00- 6.000 [^

+ 0.4+ 2.56 + 7.024
1 + 6.4 + 17.56 + 1.024

Since the value of y is positive for x = .4, the location principle
shows that (2) has a root between .3 and .4, that is, (1) has a root
between 2.3 and 2.4.

To find the root correct to two decimal places, move the origin up
to the lesser of the two numbers between which the root is now
known to lie. The new equation will have a root between and .1.

This process is performed as follows :

1 + 6.0 + 15.00 - 6.000[^-
+ 0.3 + 1.89 + 5.067

1 + 6.3 + 16.89
+ 0.3 + 1.98

1 + 6.6
+ 0.3

0.933

+ 18.87

1 + 6.9

Thus the new equation is

x^ + 6.9 x^ + 18.87 X f .933 = 0. (3)

This equation has a root between and .1. We can find an
approximate value of the hundredths place of the root by solving

THEORY OF EQUATIONS 197

the linear equation 18.87 x — .933 = 0, obtained from (3) by drop-
ping all but the term in x and the constant term.

This suggestion must be verified by synthetic division to deter-
mine between what hundredths a root of (3) actually lies.

1 + 6.90 + 18.870 - 0.9330|.04
+ 0.04 + 0.277 + 0.7658

6.94 + 19.147 - 0.1672

Thus the curve is below the X axis at a; = .04 and hence the
root is greater than .04. We must not assume that the root is
between .04 and .05 without determining that the curve is above
the X axis at x = .05.

1 + 6.90 -f 18.870 - 0.9330[^
+ 0.05+ 0.347 + 0.9608

6.95 + 19.217 + 0.0278

Thus the curve is above the X axis ai x = .05. By the location
principle (3) has a root between .04 and .05, that is, (1) has a
root between 2.34 and 2.35. We say that the root 2.34 is correct
to two decimal places. If a greater degree of precision is desired,
the process may be continued and the root found correct to any
required number of decimal places.

The foregoing process affords the following

EuLE. Plot the equation. Apply the location principle to deter-
mine between what consecutive positive integral values a root lies.

Decrease the roots of the equation hy the lesser of the two
integral values between which the root lies.

Estinfiate from the plot the nearest tenth to which the root of
the new equation lies, and determine hy synthetic division pre-
cisely the successive tenths between which the root lies.

Decrease the roots of this equation by the lesser of the two tenths
between ivhich the root lies, and estimate the root to the nearest
hundredth by solving the last two terms as a linear equation.

198

Determine hy synthetic division precisely the hundredths inter-
val in which the root must lie. ^ — ^

Proceed similarly to find the root correct to as many 'places ds
may he desired.

The sum of the integral, tenths, and hundredths values next less
than the root in the various processes is the approximate value
of the root.

To find the negative roots of an equation f(x) = 0, find the
positive roots of /(— x) = and change their signs.

When all the roots are real a check to the accuracy of the com-
putation may be found by adding the roots together. The result
should be the coefficient of the second term.

EXERCISES

Eind the values of the real roots of the following equations correct to
three decimal places.

1, x3 + 4x2 + aj + l = 0. (1)

Solution : Since by Descartes'
rule of signs there are no posi-
tive roots, we form the equation
/( — a;) = and seek its positive
root.

Thus
a;3-4x2 + aj-l = 0. (2)

Plot the equation (2) set equal
to y. In the figure two squares
are taken as the unit of x.

There is a root of this equa-
tion between 3 and 4.

Yj

<

(

X

i ^

/

\

1

>

2

k

,f

3

\

4

V

J

\

/

^

>

y
-1

-3

-7
-7
+ 3

Decrease the roots of (2) by 3,

1-4+1
+ 3-3

1[3
6

1-1 -2

+ 3 +6

1 + 2

+ 3

1 + 5

+ 4

THEORY OF EQUATIONS 199

The equation is x^ + 5x^ + Ax — 7 = 0. (3)

From the plot we estimate the root of (3) at .8.

Verify, 1 + 5.0 + 4.00 - 7.000 [^

+ 0.8 + 4.64 + 6.912
+ 5.8 + 8.64 - 0.088

1+ 5.0 + 4.00- 7.000 [^
+ 0.9 + 5.31 + 8.379
+ 5.9+9.31 + 1.379

Thus the root is determined between .8 and .9. \
Decrease the roots of (3) by .8,

1+5.0 + 4.00 -7.000^8
+ 0.8 + 4.64 + 6.912

1+5.8 + 8.64
+ 0.8 + 5.28

.088

1+6.6
+ 0.8

+ 13.92

1+7.4

The equation is x^ + 7.4 x^ + 13.92 x - .088 = 0. (4)

088

Estimate the root of (4) at x = ~ = .006.

^ ' 13.92

Verify, 1 + 7.400 + 13.920000 - .088000 [.006

+ 0.006 + 00.044436 + .083784
+ 7.406 + 13.964 - .004216
1 + 7 .400 + 13.920000 - .088000 [.007
+0 .007 + 00.051849 + .097804
+ 7.407 + 13.972 + .009804
Thus the root of the equation (1) correct to three decimal places is - 3.80(>
2. x3 - 4 = 0. 3. x* - 3 = 0.

4. X3 + X = 20. 5. 3X4 - 5X3 zz: 3I.

6. x3 + x2 = 100. 7. x3 - X - 33 = 0.

8. x4 + X - 100 = 0. 9. x3 - 8x - 24 = 0.

10. x4 - 4x3 + 12 = 0. 11. X* + x2 + X = 111.

12. x3 - x2 + X - 44 = 0. 13. x3 + lOx - 13 = 0.

14. x8 + 3 x2 - 2 X - 1 = 0. 15. x3 + x2 + X - 99 = 0.

16. x3 - 9x2 - 2 X + 101 = 0. 17. X* + x3 + x2 - 88 = 0.

18. X* - 12x3 - 16x + 41 = 0. 19. x3 - 6x2 + 5x + 11 = q.

20. x3 - 10x2 + 35x+ 50 = 0. 21. 2x4-4x3 + 3x2-1 = 0.

22. 3x4 - 2x3 - 21x2 -4x + 11 = 0. 23. 9x3 - 45x2 + 34x + 37 = 0.

200

184. Roots nearly equal. Suppose we wish to find the positije
roots, if any exist, of )

x^-{-17x^-A6x-i-29 = 0. (1)

y

4-29

+ 1

+ 13
+ 71

By Descartes* rule of signs we see that there can be
only two positive roots. We obtain the adjacent table
of values. From the plot that these values indicate we
cannot tell whether any real root exists between 1 and 2,
but if it does exist the plot indicates that it is nearer
1 than 2.

Decrease the roots of (1) by 1,

1 + 17 -46 +29[1
+ 1+18-28

1 + 18 - 28

+ 1

+ 1+19

1 + 19

- 9

+ 1

1+20

The new equation is

x^-\-20x^-

-9x-

f 1 =

0.

(2)

Estimate the root of (2) at .2 and carry the origin up to .2

and also up to .3.

1 + 20.0 - 9.00 + l.OOOj^
+ 0.2 + 4.04 - 0.992

1 4- 20.0 - 9.00 + 1.000[J
+ 0.3 + 6.09 - 0.873

1 + 20.2 -4.96
+ 0.2 + 4.08

1 + 20.4
+ 0.2

+ 0.008

1 + 20.3 - 2.91
+ 0.3 + 6.18

0.88

1 + 20.6
+ 0.3

+ 0.127

+ 3.27

1 + 20.6

1 + 20.9

By Descartes' rule of signs on the numbers obtained by moving
the origin to .3, it is seen that there are no positive roots of (2)
greater than .3, while the rule would indicate that there might
be roots greater than .2. We consider the equation

ic^ + 20.6 x^ - 0.88 X + 0.008 = 0. (3)

THEORY OF EQUATIONS 201

Estimate the roots of (3) at
_ 0.008
^ ~ 0.88

= .009.=*

Verify, 1 + 20.60 - 0.880 -h 0.00800[.01

+ 0.01 + 0.206-0.00674
1 -f 20.61 - 0.674 + 0.00126
1 + 20.60 - 0.880 + 0.00800|.02

+ 0.02 + 0.412 - 0.00936
1 + 20.62 - 0.468 - 0.00136
This determines a root of (3) between .01 and .02.

1 + 20.60 - 0.880 4- 0.00800|.03

+ 0.03 + 0.619-0.00483
1 + 20.63 - 0.161 + 0.00317

This determines another root of (3) between .02 and .03.
Decrease the roots of (3) by .01,

1 + 20.60 - 0.880 + 0.00800|.01
+ 0.01 + 0.206 - 0.00674

1 + 20.61 - 0.674
+ 0.01 +0.206

+ 0.00126

1 + 20.62
+ 0.01

- 0.468

1 + 20.63

The new equation is

x^ + 20.63 x^ - 0.468 x + 0.00126 = 0. (4)

Estimate the root of (4) at ic = ' ,^^ = .002.=*
^ ^ 0.468

Verify, 1 + 20.630 - 0.468 + 0.00126|.002

+ 0.002 + 0.041 - 0.00085
1 + 20.632 - 0.427 + 0.00041
1 + 20.630 - 0.468 + 0.00126|.003

+ 0.003 + 0.062-0.00122
1 + 20.633 - 0.406 + 0.00004

♦ We observe that in these two cases the estimated values of the roots are shown by
the verification to be inaccurate. This should insure great care in making the verifi'
cation. Th^ estimated values should nev^ b^ assumed to be accurate without verification

This indicates that a root is between .003 and .004.

Verify, 1 + 20.630 - 0.468 + 0.00126|.004

+ 0.004 + 0.083-0.00154
1 + 20.634 - 0.385 - 0.00028

This determines a root of (3) between .003 and .004.

Thus one root of (1) correct to three decimal places is 1.213.

The other root could be found similarly to be 1.229.

EXERCISES

Find all the real roots of the following equations correct to three decimal
places.

1. x^-1x-^1 = 0.

2. 7x3 -8«2_i4x + 16 = 0.

3. 2ic5_ 4x3-3x2 + 6 = 0.

4. 4x4- 5x3 -8x + 10 = 0.

5. 3x3 - 10x2 - 33x + 110 = 0.

CHAPTEE XVIII
DETERMINANTS

185. Solution of two linear equations. We have already
treated the solution of linear equations in two variables and
stated (p. 47) the method of solving three or more linear equa-
tions in three or more variables. This latter process is rather
laborious and can be very much abridged and also developed
more symmetrically by the considerations of the present chapter.
Let us solve the equations

a^x + b^ij = Ci, (1)

a^x + b^y = Ca. (2)

Multiply (1) by b^ and (2) by b^, and we obtain

ai^2^ 4- bib^i/ = b2Ci ^. v

Subtracting, we get (aib^ — a^bi) x = b^c^ — biC^,

or if aA - ^2^1 ^ 0, ^^h^LIzh^. (3)

ai/>2 — ^2^1

Similarly, we obtain y = —^ ^' (4)

We note that the denominators of the expressions for x and y
are the same. This denominator we will denote symbolically by
the following notation : ^,

aibz — a^bi = , • _

^2 t*2

The symbol in the right-hand member is called a determinant.
Since there are two rows and two columns, this determinant is
said to be of the second order. The left-hand member of the equa-
tion is called the development of the determinant. The symbols
Ui, bi, a2, ^2 ^^6 called elements of the determinant, while the ele-
ments «! and &2 are said to comprise its principal diagonal.

203

(6).

204

EuLE. The development of any determinant of the seconc
order is obtained by subtracting from the product of the ele^
ments on the principal diagonal the product of the elements on
the other diagonal.

Thus

Zi

xyi - ziy ;

12 3
4 5

10 - 12 = - 2.

Evidently each term of the development contains one and only
one element of each row and each column, that is, for instance,
the letter a and the subscript 1 appear in each term of (5) once
and only once.

We can now rewrite the solution (3) and (4) of equations (1)
and (2) in determinant form :

X =

Cl

h

ai Ci

C2

h

; y =

^2 ^2

ai

b.

a^ bi

a^

b.

a^ b^

(6)

It is noted that the numerator of the expression for x is formed
from the denominator by replacing the column which contains
the coefficients of x, a^ and a^,, by the constant terms c^ and c^.
Similarly, in the numerator of the expression for y, b^ and b^ of
the denominator are replaced by Cj and c^.

One should keep in mind that a determinant is merely a sym-
bolic form of expression for its development. In the case of
determinants of the second order the introduction of the new
notation is hardly necessary, as the development itself is simple ;
just as we should scarcely need to introduce the exponential
notation if we had to consider only the squares of numbers. It
turns out, however, as we shall see, that we are able to denote by
determinants with more than two rows and columns expressions
with whose development it would be very laborious to deal.

186. Solution of three linear equations. Let us solve the
equations

a^x + b^y 4- Ci« = c^i,
a^x 4- b^2.y + ^2^ = ^2>
<iz^ + b^y + Cg^ = c?8-

(1)

(2)
(3)

DETERMINANTS

205

Eliminating y from (1) and (2) and (1) and (3), we obtain
(ai^g — a^h^x H- (b^c^ — hiCc^)z = dj)^ — d^b^j
(a^bi — a^b^x + {c^b^ — b^c^z — d^b^ — d-J)^.

Eliminating z,

[iSHh — «2^i) (^3^1 — h(^i) — (ctsbi — a^bs) (b^Ci — ^iC2)]a;

= (dibi — d^bi) (c^bi — bsCi) — (d^bi — dj)^) (b^Ci — b^c^}

Developing, canceling, and dividing by bi, we obtain

_ d^b^Cs + d^bsCi + ^3^1 ^2 — dib^c^ — dsb^c^ — d^b^c^

aib^c^ + a^b^c^ + a^b^c^ — a^b^c^ — a^b^c^ — a^Jy^c^ ^

Following the analogy of tbe last section, we write the denomi-
nator

ai^2^3 + «2^8^1 + «8^1^2 ~" %^3<'2 "~ «3^2Cl ~ «2^1^3 =

«1

l\

Cl

^2

K

^2

^3

bs

Cs

(5)

The right-hand member of this equation we call a determinant
of the third order, and the left-hand member its development.
As in the determinant of the second order, the elements a^, b^, Cg
comprise the principal diagonal; each term of the development
contains one and only one element of each row and each column,
and all possible terms so constructed are included in the develop-
ment. The signs of the terms of the determinant of the third
order may be kept in mind by the following device.
' Rewrite the first and second columns to the right of the deter-
minant as follows :

«! ^1 G^

<Zi &i

% ^2 ^2

^2 h

«3 ^8 ^3

% h

The positive terms are found on the diagonals running down
from left to right, the negative terms on the diagonals running
up from left to right.

The numerator of the fraction (4) expressed in determinant
notation is

«! bi Cj

dz b^ Cg
dn b^ Cg

206

We can find similarly the values of y and z that satisfy equ£
tions (1), (2), and (3). The solutions of the equations in determi-
nant form are as follows :

d.

^1

^2

d.

b.

d.

h

^3

a^

K

^1

^2

h

<^2

^3

h

Oz

y =

^1

d.

Ci

^2

d.

C2

^3

d.

Cz

«1

h

^1

^2

h

^2

^3

h

Cz

«i

*i

d.

aj

*s

d.

as

Js

d>

a,

h

h

a.

h

<'2

as

h

Cs

(6)

The same principle observed on p. 204 for forming the deter-
minants in the numerators of the expressions for x and y may be
followed here. The determinant in the numerator of the expres-
sion for ic, 2/, or z is found from the denominator by replacing the
column that contains the coefficients of the variable in question
by a column consisting of constant terms. Thus in the numerator
of z we find the column d-^, d^^ d^ replacing the column Cj, Cg, Cg of
the denominator.

EXERCISES

1. Find the value of the following determinants.

3 2 1

(a)

4 6 2
1 1

•

3 2 1

Solution :

4 6 2

1 1

4

3

1

(b)

1

2

6

1

1

3

4

1

(d)

2

4

1

6

a X y

(f)

b c
c b

•

c b

(h)

-c a

-b -a

o|

= 18 + 4 + 0-6-8-0 = 8.

2

1

(c)

2

1

1

2

2

1

1

(e)

6

3

3

4

2

3

a

b

c

(g)

b

c

a

c

a

b

c

b

(i)

c

a

b

a

DETERMINANTS

207

2. Solve the following equations by determinants.

(a)

2x + 32/ = 4,

x-2y = l.
Solution : Using the expressions (6), p. 206, we obtain

-8 -3 _ n
-4-3~y

4

3

1

-2

2

3

1

-2

y =

2

4

1

1

2

3

1

-2

(e)

7 y = 12,

7 a: + 2/ = 11.

2x+5y=l, 2x + 72/:-l, a: + 4y = 2,

^ ' 7a;+6?/ = 2. ^ -* Sx-9y = 2. ^ ' 2a;-3?/ =

3. Solve the following equations by determinants.

a: + y + 2 = 2,
(a) X + 32/- 4 = 0,

y - 2 2; = 6.

Solution : Rearranging so that terms in the same variable are in a column,
and supplying the zero coefficients, we get

x+ 2/4- 2 = 2,

a: + 32/ + 02 = 4,

Ox+ y-2z = 6.

By (6), p. 206, x

2

1

1|

4

3

6

1

-2

J

1

1

1

3

1

-2

-12+0 + 4-18 + 8 +

18

1 I /-^6 + + l

0-0.+ 2

= 6,

1

2

1

1

4

6

-2

-8+0+6-0+4+0

1

1

1

-3

1

3

1

-2

1 2
3 4
1 6

1 1 1
13
1-2

Check: 6 + (-|) + (--L0) =

,18 + + 2-0-6-4
-3

6-4 = 2.

10

10

208

2ic + 3y = 12,
(b) 3a; + 2:2 = 1

Sy + 4:Z
x + y -z
(d) x + z~
V + z-

x + y +
(f) 3a:-2z =
5 y — 4 z = 0.

X + 2/ + 2 = 9,
(h) x + 2y + 3z = 14,

jc + 3y + 62; = 20.

.2x + .3y + .4z = 29,
(j) .3x + .4y + . 521=38,

.4a; + .5y + .6z = 51.

ED AMe^EBRA

7 ^lx-iy = 0,
Ac) ix-iz = l,
2; - i y = 2.
a + 2y = 7,
) 7x+92; = 29,
y+8z = 17.
X + y + 2 2; = 34,
(g) X + 2 y + 2: = 33,
2 X + y + 2: = 32.
ax + 6y — C2; = 2 a6,
(i) 6y + C2; — ax = 2 6c,
C2; + ax — 6y = 2 ac.
3x + 2y + 32: = 110,
(k) 5 X + y - 4 2; = 0,
2x-3y + 2; = 0.

187. Inversion. In order to find the development of determi-
nants with more than three rows and columns, the idea of an inver-
sion is necessary. If in a series of positive integers a greater
integer precedes a less, there is said to be an inversion. Thus in
the series 12 3 4 there is no inversion, but in the series 12 4 3
there is one inversion, since 4 precedes 3. In 1 4 2 3 there are two
inversions, as 4 precedes both 2 and 3 ; while in 1 4 3 2 there are
three inversions, since 4 precedes 2 and 3, and also 3 precedes 2.

188. Development of the determinant. In the development of
the determinant of order three we have

%^2^3 + «2^3Cl + ^S^l^a — 0^8^2Cl — ^2^1^8 " «1^3^2- (1)

the order of lettejps in each term the same as their
in the principal diagonal (as we have done in the development
above), it is observed that the subscripts in the various terms
take on all possible permutations of the three digits 1, 2, and 3.
The permutations that occur in the positive terms are 12 3,
2 3 1, 3 1 2, in which occur respectively 0, 2, and 2 inversions.
The permutations that occur in the subscripts of the negative
terms are 3 2 1, 2 1 3, 1 3 2, in which occur respectively 3, 1,
and 1 inversions.

DETERMINANTS 209

Thus in the subscripts of the positive terms an even number
of inversions occur, while in the subscripts of the negative terms
an odd number of inversions occur. This means of determining
the sign of a term of the development we shall assume in general.

When we have a determinant with 71 rows and columns it is
called a determinant of the nth order. The development of such a
determinant is defined by the following

KuLE. The development of a determinant of the nth order is
equal to the algebraic sum of the terms consisting of letters fol-
lowing each other in the same order in which they are found in
the principal diagonal hut in which the subscripts take on all
possible permutations. A term has the positive or the negative
sign according as there is an even or an odd number of inver-
sions in the subscripts.

This means of finding the development of a determinant is use-
ful in practice only when the elements of the determinant are
letters with subscripts such as in (2) below. When the elements
are numbers we shall find the value of the determinant by a more
convenient method.

In this statement it is assumed that the number of inversions
in the subscripts of the principal diagonal is zero. If this number
of inversions is not zero, the sign of any term is + or — accord-
ing as the number of inversions in its subscripts differs from the
number in the subscripts of the principal diagonal by an even or
an odd number.

Since each term contains every letter a, b, ••-jk and also every
index 1, 2, • • • , ri, one element of each row and column occurs in
each term.

In the determinant of the fourth order

tti 61 Ci di

a^ 62 C2 dz

03 &3 C3 ^i

a^ &4 C4 d^

the terms a^h^Cidi and a^h^c^di, for instance, have the minus sign, as 2 4 1 3 haa
three inversions and 4 2 3 1 has five inversions; while the terms aih^c^d^ and
a^b^Cidi have two and six inversions respectively and hence have the positive sign.

210

189. Number of terms. We apply the theorem of permutar
tions to prove the following

Theorem. A determinant of the nth order has n ! terms in
its development

Since the number of terms is the same as the number of per-
mutations of the n indices taken all at a time, the theorem follows
immediately from the corollary on p. 145.

190. Development by minors. In the development of the
determinant of order three, p. 208, we may combine the terms as
follows :

<^l (pi^^Z — h<^2) — »2 (P\<^Z — h(^l) + 0^3 (^1^2 — h<^i)

3 Cs

+ Cli

(1)

IS

We observe that the coefficient of a^ is the determinant that
we obtain by erasing the row and column in which aj lies. A
similar fact holds for the coefficients of a^ and a^. The determi-
nant obtained by erasing the row and column in which a given

element lies is called the minor of that element. Thus /

the minor of cig- ^^ notice that in the above development by
minors (1) the sign of a given term is + or — according as the
sum of the number of the row and the number of the column
of the element in that term is even or odd. Thus in the first
term a^ is in the first row and the first column, and since 1 -f- 1 = 2,
the statement just made is verified for that case. Similarly, aj is
in the first column and the second row, and since 1 + 2 = 3 is
odd, the sign is minus and the law holds here. The last term
is positive, which we should expect since a^ is in the first column
and the third row, and 1 + 3 = 4. The proof for the general
validity of this law of signs is found on p. 215.

The elements of any other row or column than the first may
be taken and the development given in terms of the minors with

DETERMINANTS

211

— (^2

Cl

+ ^'2

«1

Cl

— ^2

a^

^3

dz

<^Z

as

respect to such, elements. For instance, take the development
with respect to the elements of the second row,

d^ ^2 ^2

The rule of signs is the same as given above ; that is, for
instance, the last term is negative, as c^ is in the third column
and the second row, and 2 + 3 = 5. By generalizing these con-
siderations we may find the development of a determinant by
minors by the following

EuLE. Write in succession the elements of any row or column,
each multiplied by its minor.

Give each term a + or a — sign according as the sum of the
number of the row and the number of the column of the element
in that term is even or odd.

Develop the determinant in each term by a similar process
until the value of the development can be determined directly by
multiplication.

That this rule for development gives the same result as the definition given in
§ 188 we have seen for a determinant of order three. The fact holds in general, as
we shall prove (p. 215) .

EXERCISES

1. In the determinant of order four on p. 209 what sign should be pre-
fixed to the following terms ?

(a) Ct^hzO^di.

Solution: c^^hza-idi = azhcidi. In 2 3 4 1 there are three inversions. The
sign should be minus.

(b) a^hcsdi. (c) aibic^ds. (d) b4Cidsa2-
(e) ^8610402. (f) d2aic^b3. (g) 0208^164-

. 2. Develop by minors the following and find the value of the determinant
2 4

(a)

Solution :

1 4

1 6

3

2

3

= 3
= 3.(6

4)

|1
2(12

4) + 3 (8 - 4) = 6 - 16 + 12 = 2.

212

1 4 6

(p)

7 8 2
1 3 1
a b

(e)

d c
e f

•

2

1

(c)

2

1

3

4

a

b

(f)

a

b

a

6

(d)

5 7

3 7

-2 3

2

1
- 1

(g)

a 6
c
d e

•

(h)

Solution : Develop with respect to the elements of the first column,

2 3

3 \l

4 2
3 1

112
2 12
12 3

21

2 31

2

3

3 1 1
2 12
12 3

1 2

+ 4

3 11
112
12 3

2 3

1 1

2 3

+ 1

+ 1

: s!)-('i;

-2

1

2 3

1 1

1 2

1 1

1 2

1 2

1

+ 1

+ 2

1 2

(i)

2 6 3

i. '

= 2(-l + 2 + 0)-3(-3-2 + l) + 4(-3-l + l)-3(0-l+2)
= 2 + 12 - 12 - 3 = - 1.

1 2

8
6
2

9
1 3
1 8
1 4

ax

1 1
3
3

3 4

1 1

1 1

3 3

Hint. It is always advisable to develop with respect to the row or column that
has a maximum number of elements equal to zero.

(k)

(m)

(0)

(q)

a

b

b

a

b

a

b

a

2

3

4

1

2

3

3

6

2

3

1

1

2

3

2

1

X

a

b

b

X

a

.

a

b

X

X

y

X

y

X

y

X

y

(1)

(n)

(P)

(r)

«! 6i Ci di

62 C2 di

ai
a2 62
as

C3 f^a

^4

63
a^ 64 C4
a gr

e

a

/

a

d

c

b

X

a

b

c

X

a

b

c

X

0,

6

c

cs

a
c
b
a

X

DETERMINANTS

213

191. Multiplication by a constant. In this and the following
sections we shall give a number of theorems on determinants
which greatly facilitate their evaluation and which make a proof
for the solution in terms of determinants of any number of linear
equations in the same number of variables a simple matter.

Theorem. If every element of a row or a column is multi-
plied hy a number m, the determinant is multiplied by m.

Suppose that every element of the first row of a determinant
is multiplied by m. Since each term of the development contains
one and only one element from the first row, every term is multi-
plied by m, that is, the determinant is multiplied by m.

Illustration.

mbi bi ^3
mci Cg C3
= maiJg^a + fnazb^Ci -}- ma^biC^ — maib^c^, — ma^biC^ — ma^b^Cx

«! a

1 a.

= m

bx b^ ba

Ci C2 C3

•

6 4 1

Similarly,

8 3 2

=

10

4 1

23
24
2-5

= 2

192. Interchange of rows and columns. We now prove the

Theorem. The value of a determinant is not changed if the
columns and rows are interchanged.

Take for instance the determinant of order four.

«!

bi

Cl

di

ai

Oa

as

a^

aa

b.

C2

d.

bi

b.

bs

b.

ttg

bs

Cz

d.

Ol

^2

C3

Ci

a^

b.

C4

d.

d.

d.

d.

d.

In each of these determinants the principal diagonal is the same,
and hence the developments derived according to the statement on

214

p. 209 will be the same for each determinant, since the terms wil
be identical except for their order. The same reasoning is valic
for any determinant.

193. Interchange of rows or columns. We now prove the

Theorem. If two columns or two rows are interchanged, the
sign of the determinant is changed.

Again let us take for example the determinant of order four and
fix our attention on the first and second rows. We must prove that

«!

h

Ci

d.

a^

h.

^2

d.

as

h

Cl

d.

-

«!

W

Cl

d.

as

h

Cs

ds

as

h

Gz

ds

a^

h

Ci

d.

a^

b.

C4

d.

In the first determinant the principal diagonal is a^bc^c^d^, while
in the second the principal diagonal, a^b^^c^d^, is obtained from the
principal diagonal of the first determinant by one inversion of
subscripts. Hence this term is found among the negative terms
of the first determinant.

Since the only difference between the second determinant and
the first is the interchange of the subscripts 1 and 2, evidently
any term of the second is obtained from some term of the first
by a single inversion. Thus if a single inversion is carried out in
every term of the first determinant, we obtain the various terms
of the second. But since this process changes the sign of each
term of the first determinant (p. 213), we see that the second
determinant equals the negative of the first. Similar reasoning
may be applied to the interchange of any two consecutive rows
or columns of any determinant.

Consider now the effect of interchanging any two rows which
are separated we will say by k intermediate rows. To bring the
lower of the two rows in question to a position next below the
upper one by successive interchanges of adjacent rows, we must
make k such interchanges. To bring similarly the upper of the two
rows to the position previously occupied by the other requires k-\-l
further interchanges of adjacent rows. Hence the interchange of
the two rows is equivalent to 2 A; + 1 interchanges of adjacent

DETERMINANTS 215

rows, the effect of which is to change the sign of the determinant,
since 2 ^ + 1 is always an odd number.

194. Identical rows or columns. This leads to the important

Theokem. If a deteimiinant has two rows or two columns
identical^ its value is zero.

Suppose that the first and the second row of a determinant are
identical. Suppose that the value of the determinant is the num-
ber D. By § 193, if we interchange the first and second rows the
value of the resulting determinant is — D. But since an inter-
change of two identical rows does not change the determinant at
all, we have j) ___ t)

or 2 Z> = 0, that is, D = 0.

Corollary. If any row (or column) is m times any other
row (or column), the value of the determinant is zero.

By § 191, the determinant may be considered as the product of
m and a determinant which has two rows (or columns) identical.
Hence this product equals zero.

195. Proof for development by minors. On referring to the
rule on p. 211 we observe that in order to show that the develop-
ment by minors is the same as the development obtained by the
definition on p. 209 we must prove the two following statements.

First. The coefficient of any element in the development of a
determinant (apart from sign) is the minor of that element.

Second. Each element times its minor should have a -\- or
a — sign according as the sum of the number of the row and the
column of the element is even or odd.

Consider the element ^j.

First. Each term that contains ai must contain every other
letter than a, and the indices of these letters must take on all
permutations of the numbers 2, 3, ■ • -, n. This coefficient of ai
contains then by definition (p. 210) all the terms of its minor.

Second. Since in each term ai is in the first place, the only inver-
sions in the subscripts are those among the numbers 2, 3, • • • , ?i.

216

Hence the sign of each term in the coefficient of ai is positive or
negative according as there is an even or odd number of inversions
in its subscripts. Hence our theorem is established for the ele-
ment %.

Consider now any element, as d^ , which occurs in the fifth row
and the fourth column. Interchange adjacent rows and columns
until c?5 is brought into the leading position in the principal
diagonal. This requires in all seven interchanges, three to get
the c?6 in the first column, and then four to get it into the first
row. This changes the sign of the determinant seven times,
leaving it the negative of its original value. By the reasoning
just given in the case of % the coefficient of d^ (which is now in
the position previously occupied by ai) in the original determi-
nant would be the minor of d^, except that the signs would
all be changed. Hence the term consisting of d^ times its minor
has the — sign, and the theorem is proved for this case.

In general, to bring a term in the ith. row and kth. column to
the leading position requires i — l-{-k — l = i-\-k — 2 inter-
changes of adjacent rows or columns. This involves i -\- k — 2,
or what amounts to the same thing, i + k changes of sign. Hence
a positive or a negative sign should be given to an element times
its minor according as i + ^ is even or odd.

196. Sum of determinants. We now prove the fact that under
certain conditions the sum of two determinants may be written in
determinant form. The fact that the product of two determinants
is always a determinant is extremely important for certain more
advanced topics in mathematics, but the proof lies beyond the
scope of this chapter.

Theorem. If each of the elements of any row or any column con-
sists of the sum of two numbers, the determinant may he written
as the sum of two determinants.

For example, we have to prove that

«! + a'l

^1

Cl

«!

h

Cl

a\

^1

Cl

a^ + a\

h

C2

=

^2

h

^2

+

a'.

h.

Ci

«» -f a's

h

^3

^8

h

Cz

«'s

K

Cs

DETERMINANTS

217

Develop the first determinant by minors with, respect to the
first column, where we symbolize the minors of Ui + a'j, a^ + a'2,
ftg + a's by Ai, A^^ A^ respectively.

We have

by § 190,

«! + a'l bi Ci

^2 + a'2 ^2 ^2

as + a's ^8 ^8

«!

^1

Cl

a\

h

Cl

a^

*2

C2

+

a\

h

^2

as

^8

Cs

a\

h

Cz

= (»! -f a'O^l — (^2 + ^'2)^2 +(«8 + a'8)^8,

by the distributive law,

= aj^i — a^Az 4- 0.3^3 + a\Ay — a'2^2 + «'8^3>

by § 190,

The method of proof given is applicable to the case of any row
or column of a determinant of order n.

197. Vanishing of a determinant. For the solution of systems
of linear equations we shall make use of the

Theorem. If in the development of a determinant in terms
of the minors with respect to a certain column (or row) the ele-
ments of that column (or row) are replaced hy the elements of
another column (or row), the resulting development equals zero.

For example, we have

«!

h

Ci

d.

a?

h

Ci

d.

as

h

Cz

d.

a^

h

C4

d.

a-^A-i — a^A^ -p ^sA^ — a^A^j

(1)

where an A represents the minor of the a with the same sub-
script. We have to prove that if we replace the a's, for example,
by the b'Sj the result equals zero, that is, that

biAi — b^A^ + ^3^3 — b^A^ — 0.

(2)

218

This expression (2) when written in determinant form evidently
would have the same form as the left-hand member of (1) excej
ing that the first column would consist of ^i , ^2 ? ^3 j ^4 • ^^ should
then have two identical columns of the determinant, which would
then equal zero (§ 194). Thus the development in (2) vanishes
identically. This method of proof is perfectly general.

Corollary. The value of the determinant is unchanged if the
elements of any row (or column) are replaced hy the elements of
that row (or column) increased hy a multiple of the elements of
another row (or column).

Thus, for instance.

«!

^1

Ci

^2

h.

^2

=

as

h

Cs

«!

-i-nb.

h

Ci

^2

+ nh^

h

^2

ds

+ nh.

h

Cz

The proof follows directly from §§ 191, 196, and the preceding
theorem.

198. Evaluation by factoring. If in a determinant whose
elements are literal two rows or two columns become identical
on replacing a by h, then a — b i^ a factor of the development.
This appears immediately from § 160.

Illustration. Evaluate by factoring

(1)

Since two columns become identical if a is replaced by h, a by c,
or h by c, then we have as a factor of the development

(a-b){b-c){c-a). (2)

To determine whether the signs in this product are properly
chosen, that is, whether the development should contain a — b oy
b — a,we note that the term bc^ is positive in the development
of (1) and also positive in the expansion of (2). Evidently there
is no factor of (1) not included in (2).

1

1

1

a

b

c

a^

b'

c^

DETERMINAKTS

219

199. Practical directions for evaluating determinants. In

j&nding the value of a numerical determinant the object is to
reduce it to one in which as many as possible of the elements
of some row or column are zero. One should ask one's self the
following questions on attempting to evaluate a determinant:

First. Is any row (or column) equal to any other row (or
column) ? If so, apply § 191 for the case m = 0.

Second. Are the elements of any row (or column) multiples of
any other row (or column) ? If so, apply § 191.

Third. If we add (or subtract) a multiple of the elements of
one row (or column) to the elements of another, will an element
be zero ? If so, apply § 197.

EXERCISES

Evaluate the following determinants.

2

3

4

3

4

5

3

2

1

2

7

6

1

8

7

Solution : We obseive that if we subtract each element of the first column
from the corresponding element of the second column, the new second column
has every element 1. A similar result is obtained by subtracting the last col-
umn from the third column. Thus, by § 191,

2 3

4

3

2

1

1

4 5

3

2

4

1

1

1 2

7

6

"7

1

1

1

1

8

7

1

1

= 0.

4 3 12

6 113

4 2 12

3 6 2 1

Solution : Multiplying the last column by 2 and the whole determinant by \
does not change the value of the determinant (§ 191). Thus

4 3 12

4 3 14

6 113

_ 1

6 116

4 2 12

~ 2

4 2 14

3 6 2 1

3 6 2 2

220

Subtracting the last column from the first column and developing, we

obtam

4

3

1

4

6

1

1

6

_ 1

4

2

1

4

~2

3

6

2

2

3 14
116
2 14
16 2 2

3 14

1 1 6

2 14

Subtracting the last row from the first row,

3 1

1 1

2 1

1 6
1 4

-,-(-2) = l.

a — d a 1
b-d b 1
c — d c 1

a^ + b
2ab

2ab

1

a2 + 62

1

2a6 + 62

1

5.

7.

9.

3

3

4

2

1

1

2

1

2

2

3

1

2

1

3

2

a;

y

2

X

y

2

y

2

«

2

y

a;

a

1

&

1

1

c

1

2

d

1

3

3

10.

4

6

8

3

1

1

2

1

2

3

4

1

2

1

3

4

110 1
10 11
111

1111

1

1

1

1

1

1

1

1

1

1

1

1

11.

a

a

a

a

a

b

b

b

a

b

c

c

a

b

c

d

12.

2

1

1

1

1

2

1

1

1

1

2

1

1

1

1

2

13.

3

7

16

14

6

15

33

29

1

1

1

4

2

3

1

14.

8

2

1

4

16

29

2

14

16

19

3

17

33

39

8

38

15.

9

13

17

4

18

28

33

8

30

40

54

13

24

37

46

11

16.

[2

14

16

18

2

4

6

8

4

3

2

1

3

7

11

16

DETERMINANTS

221

D,

200. Solution of linear equations. Suppose that we have given
n linear equations in n variables. We seek a solution of the equa-
tions in terms of determinants. For simplicity, let n — 4:. Given

a^x + hy + Ci« + d^w =/i, (1)

a^x -f h^y + c^z + d^w =/2, (2)

(^3^ + hy + CsZ + d^w =/8, (3)

a^x + hy + C4^ + d^w =/4. (4)

The coefficients of the variables taken in the order in which
they are written^ift^ be taken as forming a determinant D, which
we call the determinant of the system. Thus
% hi Ci di

(^2 t>2 ^2 ftg
^8 ^3 ^3 ^8

a^ b^ C4 d^

Symbolize by ^1, ^3, etc., the minors of %, 63, etc., in this deter-
minant. Let us solve for x. Multiply (1), (2), (3), (4) by ^1, A^,
Azj A^ respectively. We obtain

A^a^x + A^biy + A^c^z + A^d^w = ^j/i,
A^a^x + A^b^^y + ^3^2^ + A^d^w = ^2/2,
^gaga: + A^b^y + ^3^3^ + ^af/gi^ = ^3/3,
A^a^x -f- ^4^'4?/ 4- A^c^z + ^4C?4i^; = A^f^.

If we add these equations, having changed the signs of the
second and fourth, the coefficient of x is the determinant Z>, while
the coefficients of y, z^ w are zero (§ 197). The right-hand mem-
ber of the equation is the determinant D, excepting that the ele-
ments of the first column are replaced by fufi, fs,f^ respectively.
Hence

X =

/l

h

Cl

d.

A

h

Ca

d.

A

h

Cs

d.

A

h

^4

d.

«!

h

Cl

d.

^2

h

C2

d.

^8

h

Cz

d.

a^

h

Ci

d.

222

In a similar manner we can show that the value of any variable
which satisfies the equations is given by the following

EuLE. The value of one of the variables in the solution of n
linear equations in n variables consists of a fraction whose
denominator is the determinant of the system and whose numer-
ator is the same determinant, excepting that the column which con-
tains the coefficients of the given variable is replaced by a column
consisting of the constant terms.

When D = 0, we cannot solve the equations unless the numer-
ators in the expressions for the solution also vanish.
Illustration. Solve for x

ax -\-2by — 1,

2by + ^cz = 2,

S cz -{- 4: dw = Sj

4 dw -{- 5ax = A.

E-earranging, we obtain

ax + 2 by = ly

2by-^3ez =2,

3 cz -{- 4: dw = 3j

-{- 4:dW = 4:.

D

5 ax

a 2b

2b 3c

3c

5a

v"

= 24

4:d

a 5

5 c

c ^

5a ^ ^ d

1

b

1

b

24

2

b

c

1

c

3

c

d

3

c d

X —

4

d

4

d

a

b

a

b

24

b

c

b c

c

d

c d

5a

d

-5b d

1

c

-b

3

c

d

4

d

b

c

a

c

d

-5b

d

DETERMINANTS

223

1

c

-h

2

d

4

d

b

G

a

C

d

5c

d

be

ab

G d
5c d

- 2 bed
— 4 abed

1_

2a

201. Solution of homogeneous linear equations. The equa-
tions considered in the previous section become homogeneous
(p. 115) if /i =/2 =/3 =/, = 0. We have then

a^x + b-^y -\- GiZ + diW = 0,
azX -f bzy + G2% + dzW = 0,
^3^5 + b^y + ^3^; + cZgW; = 0,

(I)

These equations have evidently the solution x = y = z = w =0.
This we call the zero solution. We seek the condition that the
coefficients must fulfill in order that other solutions also may
exist. If we carry out the method of the previous section, we
observe that the determinant equals zero in the numerator of
every fraction which affords the value of one of the variables
(§ 191). Thus if D is not equal to zero, the only solution
of the above equations is the zero solution. This gives us
the following

Pkinciple. a system of n linear homogeneous equations in n
variables has a solution distinct from the zero solution only
when the determinant of the system vanishes.

Whether a solution distinct from the zero solution always
exists when the determinant of the system equals zero we shall
not determine, as a complete discussion of the question would be
beyond the scope of this chapter.

Theorem. If x^, y^, z^, w^ is a solution of equations (I)
and h is any number, then kx^, ky^, kz^, kw^ is also a solution.

The proof of this theorem is evident on substituting kx^ etc.,
in equations (I) and observing that the number A; is a factor of
each equation. Thus if a system of n linear homogeneous equar
tions has any solution distinct from the zero solution it has an
infinite number of solutions.

EXERCISES
Solve for all the variables :

x + y = a, a; + 3 y = 19,

y + z = b, y-]-Bz = S,

M + « = d, w + 3d = 11,

V + x = e. V + 3ic = 15.

z-\-y + w = a, Bx + y + z = 20,

^z + w + x = b, -a; + 4y + 3w; = 30,

*w + a; + y = c, '6jc + 2 + 3iy = 40,

x-{-y-\-z = d. 8y + 3z + 6w; = 50.

2/ + 2; + 5iy = ll, x — 2y + Sz — w = 6,

^ x + z-\- 4:W = llj ^y-2z + Sw-x = 0^

'a: + y + 3M; = ll, ' z-2w-\-Sx-y = 0,

x + z + Sy = 3S. w-2x + Sy - z = 6.

7.

x + y-\-z + w = 2i, x + y + z + w = 60,

a; + 2y + 3z-9w = 0, g x + 2y + 3z -\- ^w = 100,

3x-y -6z-\-w = 0, ' x-\-Sy -{■Qz-\-10w = 150,

2x + 3y--4«- 610 = 0. x + 42/ + lOz + 20mj = 210.

CHAPTEE XIX
PARTIAL FRACTIONS

202. Introduction. For various purposes it is convenient to

f(x)
express a rational algebraic expression ~j-\ ? § 11, as the sum of

several fractions called partial fractions, which have the several
factors of <^{x) as denominators and which have constants for
numerators. If we write <j>(x) = {x — a)(x — P)--- (x — v), we
seek a means of determining constants Aj B, ■, N such that for
every value of x

<l>{x) X — a X — p X — V ^ ^

If the degree of f(x) is equal to or greater than that of ^ (a;),
we can write

<i>(x) <}>(x) ^ ^

where Q (x) is the quotient and f^ (x) the remainder from dividing
f(x) by <f>(x), and where the degree oi /^(x) is less than that of
<^ (x). In what follows we shall assume that the degree of f(x)
is less than that of <f> (x). In problems where this is not the case
one should carry out the long division indicated by (2) and apply
the principle developed in this chapter to the expression corre-
sponding to the last term in (2).

203. Development when </>(x) = has no multiple roots. Let
us consider the particular case

f(x) ^ x + 1

<t> (x) (x -l)(x- 2) (x - 3)'

We indicate the development required in form (1) of the last
section, . . « ^

(x-l)(x-2)(x-3) x-1 x-2 x-S
225

where A, B, and C represent constants which we are to deter-
mine if possible. The question arises immediately, Are we at
liberty to make this assumption? Are we not assuming the
essence of what'we wish to prove, i.e. the form of the expansion?
To this we may answer. We have written the expansion in form (1)
tentatively. We have not proved it and are not certain of its
validity. If, however, we are able to find numerical values of
Ay B, and C which satisfy (1), we can then write down the actual
development of the fraction in the form of an identity.

If, on the other hand, we can show that no such numbers A,BjC
satisfying (1) exist, then the development is not possible.

Clear (1) of fractions,

X + 1 = A(x - 2){x - ^)+ B{x -l){x - ^)-\- C (x -l)(x - 2)
= {A + B-^C)x^-{6A +4.B-\-^C)x^QA+^B-\-2C.

Since we seek values of A, B, and C for which (1) is identi-
cally true for all values of x, equate coefficients of like powers of
X in the last equation (Corollary II, p. 174). We obtain

(2)
(3)
(4)

(5)
(2)

A-\-B+ C =

-5A _45-3C=

6A +SB-\-2C =

= 0,
-.1.

to (3) and

we obtain

A -B- C=

A+B-hC =

2A =

A =

= 2,

:1 ^

From (^]

1 and (4) we obtain

B=-3j C =

:2.

Tlma

x

+ 1

1

XIlUo

(x-l)(x

-2)(a.-3)

X-1

4-

As a check we might clear of fractions and simplify. If equar
tions (2), (3), and (4) had been incompatible, we should have con-
cluded that we could not develop the fraction in form (1).

PARTIAL FRACTIONS 227

We assume now for the general case

^(x) = (x - a)(x - P)--(x - v),

and that the n roots a, )3, • • •, v are all distinct from each other.
Let us consider the expression

f(x)_ A B N

<^{x) X — a X — p X — V ^

where A, B, •", N are constants. Let us assume for the moment
the possibility of expressing ~y^ in terms of these partial frac-
tions. We shall now attempt to determine actual values A, B,--,
iV which satisfy such an identity. K we multiply both sides of
the identity by

<t>(x) = (x-a)(x-P)-'(x-v),
we obtain

f(x) = A(x-P)'"(x-v) + B(x-a)--'(x-v) + -"

^N(x-a)(x-P)";

f(x) is of degree not greater than n — 1, and consequently when
written in the form of (1), p. 166, has not more than n terms. If
we multiply out the righ1>hand member and collect powers of ar,
we have an expression in a; of degree n — 1. By Corollary II,
p. 174, this equation will be an identity if we can determine
values ot Aj Bf "-,N which make the coefficients of x cm both
sides of the equation equal to each other. Hence we equate
coefficients of like jx>wers of x and obtain n equations linear in
A, B, '",N which we can treat as variables. These equations have
in general one and only one solution which we can easily deter-
mine. The values of Aj B, • • • , iV obtained by solving these equa-
tions we can substitute for the numerators of the partial fractions
in (6). After making this substitution we can actually clear of •
fractions the right-hand member of (6) and check our work by
showing its identity with the left-hand member.

There is no general criterion that we have applied to (6) to
determine whether the n linear equations obtained lay equating
coefficients of x have any solution or not. Hence in this general

discussion it should be distinctly understood that assumption (6)
holds when and only when these equations are solvable. In
any particular case we can find out immediately whether the
equations are solvable by attempting to solve them. If the num-
bers Aj By ' • ', N do not exist, the fact will appear by our inability
to solve the linear equations. As a matter of fact, one and only
one solution always exists under the assumption of this section.

If in (6) we assume that several of the symbols A, B, ■'■, N stand for expres-
sions linear in" x, as, for instance, ax + b, we should then have a larger number of
variables to determine than there are equations. Under these circumstances there
is an infinite number of solutions of the equations. Thus if we should seek to

express =^-7— as the sum of partial fractions where the numerators are not con-

stants but functions of z, we could get any number of such developments.

We have the following

EtTLE. Factor <\>{x) into linear factors, as
(x — a)(x — P)"-{x — v).
Write the expression

</)(a?) X — a X — P x — v

Multiply both sides of the expression by <i>{x)y equate coefficients
of like powers of x, and solve the resulting linear equations for
A,B,-^-,N.

Replace A, B, •", N hy these values and check hy substituting
for X some number distinct from a, ^, "•, v.

EXERCISES

Separate into partial fractions ;

(x-l)(x-2)x

/J.2 _ 2 A B f*

Solution : Assume = 1 1 — . (1^

{x-l){x-2)x a;-lx-2a; ^'

Multiply by (x - 1) (x - 2) x,

x2 - 2 = ^(x - 2)x + J5(x - l)x + C(x - l)(x - 2),

3fi-2 = {A+B-^C)x^-{2A+B + SC)z -f 2^.

PARTIAL FKACTIOKS 229

Equating coefficients of like powers of x,
A + B + C = l,
2A+B + SC = 0,
2 O = - 2.
Hence C = — 1.

Solving

we obtain

Thus

A + Bz

= 2,

2A + B =

= 3,

A =

= 1

J5 =

= 1

■ ' +-

1

x^-2 1.1 1

(a;-l)(x-2)a; z-1 x-2 x
Check : Let

Substituting in (1), — - = — ~ -\

x =

-A,

-1

-6

r

-■2

*^.-

1_
6~

1
2

-b'

5+1=1.

6 -6

2.

x-1

x^ -\-Sx-^2

A

1

3 a;2 _ 2 X - 8

(i

4a;2

{x^-i){x-l)

R

a;2_3a; + l

3. " + ''

1.

2a;2-5x-3

5x

6x2-5x-l

2x2-1

(x2 + 3a; + 2)(x-

-1)

a;2 4-4

9.

(X - 1) (X - 2) (X - 3) (X - 2) (X + 2) (X - 1)

204. Development when ^ (a?) = O has imaginary roots. In

the preceding section no mention has been made of any distinc-
tion between real and imaginary values oi a, p, ■ -- , v. In fact
the method given is valid whether they are real or imaginary. It
is, however, desirable to obtain a development in which only real
numbers appear.

Let us assume the development

iS=-^ + -^ + - + ^^ + -^' (1)

^ {x) X — a X — p X — fji X — V

where let us suppose that /x and v are the only pair of conjugate
imaginary roots of <^ (.r), m and n being conjugate complex numbers.

Let fi = a-\-ib, v = a — ib.

Then adding the corresponding terms of (1), we obtain

m n _ X (m -{- n) — a (m + n) b(m — n)

x — a — ib x — a + ib (x — a)^ -{- b^ (x — af-\-b'^ ^ ^

Since /a and y are the only imaginary roots of <f> (x) = 0, the last
term of (2) is real, as is also the entire right-hand member (§ 152).
Hence, letting the numerator

x(7}i-\-n)— a(m-\-n)-\- ib (m — n) — Mx + N,

we have the development

/M^^4_ _B_ ■ Mx + N

<^(x) x-a x- ft "^ (ic _ a)2 -|. J2 W

By complete induction we can establish this form of numerator
where there is any number of pairs of imaginary roots of <f>(x) = 0.
^e have proved the form (3) where there is one pair of imagi-
nary roots. Assuming the form where there are k pairs, we can
prove it similarly where there are A: + 1 pairs. Hence we have the

Theorem. If <l>(x) is facto7^ahle into distinct linear and quad-
ratic factors, but the quadratic factors are not further reducible

fix)
into real * factors, then is separable into partial fractions

of the form ^^ ^

A B Mx + N

+ ;;+•••+ o ■ ■ >

X — a X — P a? -\- ^x + V

where x^ + fix -\- v is an irreducible quadratic factor of <t>(x).

This theorem is of course true only under the condition that
the linear equations obtained in the process of determining the
constants are solvable. It turns out, however, that in this case
as in § 203 the linear equations obtained always have one and
only one solution provided that the roots are all distinct.

* A real /actor is one whose coefficients are all real numbers.

PARTIAL FRACTIONS

231

EXERCISES

Separate into partial fractions :

x2+ 1

(X - 1) (x2 + X + 1)
Solution : Assi^me

X2+1

A Bx + C

...^ (X - 1) {X2 + X + 1) X - 1 X2 + X + 1

Multiply by (x -^{af^ + a; + Ij, - '

x2 + 1 = ^ (x2 + x + 1) + (Bx + 0) (« - 1).

Collecting like powers of x,

3fi + l = {A + B)x^-{- {A- B+ C)x + A-C.

Equating coefficients of x,

A + B = l,

A-B-{-C = 0,

A-C = l.

Add (2) and (3) and solve with (1),

2A-B=1

A-{-B=:l

Substituting in (1),
Substituting in (3),

SA = 2,

A = l

B = h

>

(1)
(2)
(3)

Thus

X2 + 1

x-1

(x-l)(x2 + x + l) 3(x-l) 3x2 + 3x + 3

Check: Let x = — 1.

2 2 -2

Substituting,
Reducing,

„ X2 + a + 1

2-1 3

6 3

2 3-3+3
1.
x2 + l

x3 + 4x

x2 + 4

x8-2x2 + 3x-

X

2

(X + 3) (2 x2 - X

X2 + 1

-4)

X* + X2 + 1

5x^-1

x4 + 6x2 + 8*

1

(X - l)(3x2 + x + 6)

x3 + 3x2-2x-16
Hint. Factor by synthetic divi-
sion (see p. 178) .

205. Development when ^ (x) = (x — a)~. In this case the
method given in the previous sections fails, as the equations for
determining the values of the numerators are incompatible. If

^®^®* f{x) ^ a,x-' + a,x-' +... + a^_^

<f>(x) (x-ay ' ^ ^

we can separate into partial fractions as follows.

Let X — a = y, that is, x = y -\- a, and substitute in (1). We
obtain after collecting powers of y

_ = 1 — -^ f_ — _,

y y y y

where the ^'s are constants. Eeplacing yhyx — a,wQ have th^
following development :

A,

/(f) = ^0 I ^1 I ^'^ I ....).

(x — a)" X — a (x — ay (x — ay (x — ay

EXERCISES

Separate into partial fractions :

- x2 + 2 a; - 1

(X-3)8

Solution : Assume — -— = H — ■ -(- — -. (1)

(X - 3)8 jc - 3 (X - 3)2 (X - 3)8 ^ '

Multiply by (x - 3)8,

x2 4- 2 X - 1 = ^ (x2 - 6x + 9) + 5(x - 3) + C

Collecting powers of x,

= Ax'^ + {B-6A)x-\-9A-BB-\-C.
Equating coefficients of x,

^ = 1,

5-6^ = 2,

9A-SB + C = -1.

Solving, B = 8, C = U.

„ x2 + 2x-l 1,8 14

Hence — ' = 1 \-

(X - 3)8 X - 3 (X - 3)2 (X - 3)8

Check : Let x = 1.

^ ^ ,.,,.. ,,, 2 1 8 14 1 „ 14

Substituting in (1), __ = — - + --_=_- + 2-— ,

1 1

PARTIAL FRACTIONS 233

2. ^^^.. 3.

(x-2)3 (2a; + 1)2 (cc - 4)3

g x^ + x + 1 g a;- g - 2 a;^ + 3 x + 1

(2x-l)*" ' (ax + 6)2* * (3x-2)3

206, General case. When <f>(x)= has real, complex, and
multiple roots, we may use all the previous methods simultane-
ously. Hence for this case we assume the expansion

f(^

(x — a)--- (Xx^+ fxx -\- v) ■ " (x — tY

X — a Ax2-f fjiX -\- V X — T (£c — t)'

EXERCISES

Separate into partial functions :
- g^ + 2x2 + i8x-18
' (x-l)(x2 + x + l)(x-2)2'

Solution :

x* + 2x2 + 18x-18 A Bx-VG D . E

(X - 1) (x2 + X + 1) ^ - 2)2 X - 1 x2 + X + 1 X - 2 (X - 2)2

= ^(x2 + X + 1) (x - 2)2 + {Bx + C) (X - 2)2 (X - 1)
+ i)(x-l)(x-2)(x2 + x + l)

+ JSJ(X-1)(X2 + X + 1)

= {A-hB+D)x^-\-{-3A-6B + C-2D-\-E)x^
-{-{A + 8B-5C)x2 + (-4E + 8C'-D)x
+ (4 ^ - 4 C + 2 D - ^).

Equating coefficients of like powers of x,

A+ B + D =1, (1)

-SA-6B-\- C-2D + E = 0, (2)

A-^8B-5C =2, (3)

- 4 B + 8 C - D =18, (4)

4 A -4C + 2D-^ = -18. (6)

Adding (2) and (5), (1) and (4), we have, together with (3),

^_5E-3C' = -18, (6)

A-3B-\-SC = 19, (7)

A-\-SB-6C = 2. (8)

Adding all three equations, we find

3^ = 3, or A = l.

Substituting in (3) and (7) and solving, we find C = S, B = 2. Substituting
in (1), we find D = - 2. Similarly, from (6), ^ = 6.

x* + 2a;2 + 18x-18 1 , 2a; + 3 2 , 6
Thus — — - = + — 4-

(x-l)(x2 + a; + l)(x-2)2 x-1 x^ + x + 1 x-2 (a; -2)2

Check: Let x = — 1.

go J 2 2 6

Substituting, -^-^ = __ + ____ + _,

¥ = 2i-i = -V-.
2. -^!±^. 3.

X{X- 1)3

. X3 -f 2 X2 + 1 _

4. o.

X(X-1)3

g 2x3 + x + 3 y

(X2 + 1)2

8 ^~^ 9

■ (x + l)2(x + 2)'

10. ^^^ + ^ . 11.

(a:2-l)(x + 2)

12. ^^-^^ + ^ . 13.

(x - 8) (X - 9)

.. 2x2 -3x- 3 .-

14. 15.

16. "^^ — tl 17.

(2x-3)(6x2-6x + l) (a;2_3a;4.2)(x-3)

l-x*

5X + 12

x(x2 + 4)

43X-11

30(x2-l)

X3 - X - 1

X4-16

x2 + 6x-8

x3-4x

2

(x2 + X + 3) (2 a

• + 1)

x3 4-2x2-3x

+ 1

(x + 3)(x2-4x

+ 5)

13 -5x

CHAPTEE XX
LOGARITHMS

207. Generalized powers. If h and c are integers, we can easily
compute h''. When c is not an integer but a fraction we can com-
pute the value of ¥ to any desired degree of accuracy. Thus if
6 = 2, c = I, we have 2^ = V2* = VS, which we can find to any
number of decimal places. If, however, the exponent is an irrar
tional number as V2, we have shown no method of computing
the expression. Since, however, V2 was seen (p. ^5) to be the
limit approached by the sequence of numbers

1, 1.4, 1.41, 1.414, ••,

it turns out that 5^^ is the limit approached by the numbers

The computation of such a number as S^"*^ would be somewhat
laborious, but could be performed, since 5^-^^ = 5**^ = V5^*\ Thus
it is a root of the equation x^^ = S^'*^ and could be found by
Horner's method, p. 197.

We see in this particular case that 5^^ is the limit approached
by a sequence of numbers where the exponents are the successive
approximations to V2 obtained by the process of extracting the
square root. In a similar manner we could express the meaning
of 5", where i is a positive integer and c is any irrational number.

Assumption. We assume that the laws of operation which we
have adopted for rational exponents hold when the exponents are
irrational.

Thus 6c.6d= 6c+d^ ^= 6e-d, {Tbd)c= {ho)d= bcd^

where c and d are any numbers, rational or irrational.

236

208. Logarithms. We have just seen that when h and c are
given a number a exists such that ¥ = a. We now consider the
case where a and h are given and c remains to be found. Let
a = 8, ^ = 2. Then if 2*^ = 8, we see immediately that c = 3 satis-
fies this equation. If a = 16, 5 = 2, then 2" = 16 and c = 4 is the
solution. If a = 10, 6 = 2, consider the equation 2" = 10. If we
let c = 3, we see that 2^ = 8. If we let c equal the next larger
integer, 4, we see 2* = 16. If then any number c exists such that
2*^ = 10, it must evidently lie between 3 andj:. To prove the exist-
ence of such a number is beyond the^^ope of this chapter, but
we make the following

Assumption. There always exists a real number x which
satisfies the equation

b^ = a, (1)

where a and h are positive numbers, provided b ^ 1.

Since any real number is expressible approximately in terms of
a decimal fraction, this number x is so expressible.

The power to which a given number called the base must be
raised to equal a second number is called the logarithm of the
second number.

In (1) X is the logarithm of a for the base b.

This is abbreviated into

X = \ogf,a. (2)

Thus since

28 = 8, 102 = 100, 3- 2 = J, 40 = 1,
we have

3 = logaS, 2 = logiolOO, - 2 = logg j, = log4l.

The number a in (1) and (2) is called the antilogarithm.

EXERCISES

1. In the following name the base, the logarithm, and the antilogarithm,
and write in form (2).

(a) 36 = 729.

Solution : 3 = base, 6 = logarithm, 729 = antilogarithm, logs 729 = 6.

(b) 2* = 16. (c) 38 = 27.

LOGARITHMS 237

2. Find the logarithms of the following numbers for the base 3 :

81, 243, 1, i, ij.

3. For base 2 find logarithms of 8, 128, |, ^V

4. What must the base be when the following equations are true ?
(a) log 49 = 2. (b) log81 = 4.

(c) log 225 = 2 . (d) log 625 = 4.

209. Operations on logarithms. By means of the law expressed
in the Assumption, § 207, we arrive at principles that have made
the use of logarithms the most helpful aid in computations that
is known.

Theorem I. The logarithm of the prodicct of two numbers is
the sum of their logarithms.

Let log^a. = a;,

logfeC = y.
Then by (1) and (2), p. 236, h^ = a,

¥ = c.
Multiply (by Assumption, § 207),

or by (1) and (2), \og^a-c = x -\- y.

Theorem IL The logarithm of the nth power of a number is
n times the logarithm of the number.

Let

logi,a = Xy

or

b' = a.

Raise both sides to the nth. power,

(b^f. = b'^ = a%

or

logft a" = nx.

Example.

Iogiol00 = 2,

logio 1000 = 3.

By Theorem I,

logio 100,000 = 5,

which is evidently true,

since 10^ = 100,000.

Theorem III. The logarithm of the quotient of two
is the difference between the logarithms of the numbers.

Let logj, a = Xj

log6C = 2/.
Then b'' = a,

b^ = c.

Dividing, ^,«-y = -,

c ^^

, a
or log;, -==x — y.

G

Theorem IV. The logarithm of the real nth root of a nur.t-
ber is the logarithm of the number divided by n.

Let logj, a = Xj

or b"" = a.

1 X

Extract the nth root, (b'^y = ^^ = -Va,

or logj -Va = - •

n

EXERCISES

Given logio2 = .301, logio5 = .699, logio? = .8451, find

1. log(-s/f^- V5).*
Solution :

By Theorem I, log ( v^T^ • VS) = log v^ + log V5.
By Theorems III and IV, = f log 7 + ^ log 5.

Now f log 7 = f (. 8451) = . 50706,

and i log 5 = |_(.699) = .349^

Adding, | log 7 + | log 5 = log ( Vt^ . VB) = .85656

2. log 40. 3. log 28.
Hint. Let 40= 8-5= 28.6.

4. log 140. 5. log V280.

6. log V'35. 7. log ( V8 . v^ . V^).

8. log ( V5 . 78). 9. log ("t/Ie . Vli . 4^700),

• Where no base is written it is assumed that the base 10 is employed.

LOGARITHMS 239

210. Common system of logarithms. For purposes of compu-
tation 10 is taken as a base, and unless some other base is indi-
cated we shall assume that such is the case for the rest of this
chapter. We may write as follows the equations which show the
numbers of which integers are the logarithms.

Since 10^ = 100,000 we have log 100,000 =A

10* = 10,000

log 10,000

= 4.

108 ^ IQQQ

log 1000

= 3.

102 ^ IQQ

log 100

= 2.

10^ = 10

log 10

= 1.

10« = 1

logl

= 0.

io-» = .i

log.l

= -1.

10-2 = .01

log .01

= -2.

10- « = .001

log .001

= -3.

etc.

etc,

Assuming that as x becomes greater log x also becomes greater,
we see that a number, for example, between 10 and 100 has a
logarithm between 1 and 2. In fact the logarithm of any number
not an exact power of 10 consists of a whole-number part and a
decimal part.

Thus since IQS < 3421 < lO^,

log 3421 = 3. + a decimal.
Since 10- 3 < .0023 < 10-2,

log .0023 = - 3. + a decimal.

The whole-number part of the logarithm of a number is called
the characteristic of the logarithm.

The decimal part of the logarithm of a number is called the
mantissa of the logarithm.

The characteristic of the logarithm of any number may be seen
from the above table, from which the following rules are imme-
diately deduced.

'f The characteristic of the logarithm of a number greater than unity
is one less than the number of digits to the left of its decimal point.

Thus the characteristic of the logarithm of 471 is 2, since 471 is between 100
and 1000; of 27.93 is 1, since this number is between 10 and 100; of 8964.2 is 3,
since this number is between 1000 and 10,000.

The characteristic of the logarithm of a number less than 1
is one greater negatively than the nwmher of zeros ^preceding the
first significant figure.

Thus the characteristic of the logarithm of .04 is — 2 ; of .006791 is — 3 ; of
.4791 is - 1.

It must constantly be kept in mind that the logarithm of a
number less than 1 consists of a negative integer as a character-
istic plus a positive mantissa. To avoid complication it is desir-
able always to add 10 to and subtract 10 from a logarithm when
the characteristic is negative. Thus, for instance, instead of writ-
ing the logarithm - 3 + .4672 we write 10 - 3 + -4672 - 10, or
7.4672 — 10. This is convenient when for example we wish to
divide a logarithm by 2, as by Theorem IV, § 209, we shall wish
to do when we extract a square root. Since in the logarithm
— 3 -f .4672 the mantissa is positive, it would not be correct to
divide — 3.4672 by 2, as we should confuse the positive and
negative parts. This confusion is avoided if we use the form
7.4672 - 10, and the result of division by 2 is 3.7336 - 5, or
8.7336 — 10. The actual logarithm which is the result of this
division is — 2 + .7336.

Theorem. Numhers with the same significant figures which
differ only in the position of their decimal points, have the same
mantissa.

Consider for example the numbers 24.31 and 2431.

Let 10^ = 24.31.

. Then a = log 24.31.

If we multiply both numbers of this equation by 100, we have
10^10^ = 10^+2^2431,
or x-\-2 = log 2431.

Thus the logarithm of one number differs from that of the other
merely in the characteristic. In general numbers with the same sig-
nificant figures are identical except for multiples of 10. Hence their
logarithms differ only by integers, leaving their mantissas the same.

Thus if log47120. = 4.6732, log47.12 = 1.6732, and log .004712 = - 3.6732, or
7.6732 - 10.

LOGARITHMS 241

EXERCISES

By §209,

log V600 = 1.38905

2. log .06.

3. log (210)3.

4. log

5. log(4.2)4.

6. log

"^2.1

7. log

a , 567

Q , 13.23

9. log

^ 1.28

If log 2 = .3010, log 3 = .4771, log7 = .8451, find
1. logVOOO.

Solution : log V600 = log V20 • 30 = ^ log 20 + i log 30.

By the preceding theorem, log 20 = 1.3010, log 30 = 1.4771.
i log 20= .6505
I log 30= .73855

(70)3
324

211. Use of tables. A table of logarithms contains the man-
tissas of the logarithms of all numbers of a certain number of
significant figures. The table found later in this chapter gives
immediately the mantissas for all numbers of three significant
figures. In the next section a method is given for finding the
mantissa for a number of four figures.* Hence the table is called
a four-place table. Before every mantissa in the table a decimal
point is assumed to stand, but in order to save space it is not
written. To find the logarithm of a number of three or fewer
significant figures we apply the following

Rule. Determine the characteristic hy rules in § 210.

Find in column N the first two significant figures of the num-
ber. The mantissa required is in the row with these figures.

Find at the top of the page the last figure of the number.
The mantissa required is in the column with this figure.

When the first significant figure is 1 we may find the loga-
rithm of any number of four figures by this rule from the table
on pp. 248, 249 if we find the first three instead of the first two
figures in column N.

Thus the log 516. = 2.7126, log .00281 = - 3.4487, log 7400. = 3.8692,
log 600. = 2.7782, log 50. = 1.6990, log 4.00 = .6021.

EXERCISES
Find the logarithms of the following :

1. 3. 2. 303. 3. .024.

4. 347. 6. .0333. 6. 1.011.

7. .202. 8. .0029. 9. .0001.

10. .00299. 11. 68400. 12. .0201.

212. Interpolation. We find by the preceding rule that,
log 2440 = 3.3874, while log 2450 = 3.3892. If we seek the loga-
rithm of a number between 2440 and 2450, say that of 2445,
it would clearly be between 3.3874 and 3.3892. Since 2445 is just
halfway between 2440 and 2450, we assume that its logarithm is
halfway between the two logarithms. To find log 2445, then, we
look up log 2440 and log 2450, take half (or .5) their difference,
and add this to the log 2440. This gives

log 2445 = 3.3874 + .5 x .0018 = 3.3883.
If we had to find log 2442 we should take not half the difference but
two tenths of the difference between the logarithms of 2440 and
2450, since 2442 is not halfway between them but two tenths of the
way. This method is perfectly general, and we may always find the
logarithm of a number of more than three figures by the following

Rule. Annex to the proper characteristic the mantissa of
the first three significant figures.

Multiply the difference between this mantissa and the next
larger mantissa in the table (called the tabular difference and
denoted by D) by the remaining figures of the number preceded
by a decimal point.

Add this product to the extreme right of the logarithm of the

first three figures, rejecting all decimal places beyond the fourth.

In this process of interpolation we have assumed and used the principle that
the increase of the logarithm is proportional to the increase of the number. This
principle is not strictly true, though for numbers whose first significant figure
is greater than 1 the error is so small as not to appear in the fourth place of the
logarithm. For numbers whose first significant figure is less than 2 this error
would often appear if we found the fourth place by interpolation. For this reason
the table on pp. 248, 249 gives the logarithms of all such numbers exact to four
figures, and in this part of the table we do not need to interpolate at all

LOGARITHMS 243

EXERCISES

Find the logarithms of the following :
1. 63.924.

Solution: log 63.9 = 1.8055

2

Tabular difference =

4. 62230.

7. 20060.

10. 9.999.

13. 5.7828.

16. 3.1416.

log 275 =2.4393

D= 16

6
2.4399

.4
6.4

log63.924 = 1.8057 iL

1.68
We add 2 to 1.8055 rather than 1, because 1.68 is nearer 2 than 1. In general
we take the nearest integer.

2. 269.4. 3. 1001.

5. 392.8. 6. 9.365.

8. .4283. 9. .3101.

11. 82.93. 12. .05273.

14. .003011. 15. .002156.

17. 276.4 X 1.463.
Solution : log 275.4 = 2.4399

log 1.463 = 0^652
By Theorem II, § 209, log (275.4 x 1.463) = 2.6051

18. 374.3 X 1396. 19. 1.46 x 237.2.
20. 469.1 X 63.92. 21. 47320. x .8994.

22. :5?!?1.

^8-^^ log .0372 =8.5705 -10 /)= 12

Solution: log. 03724 = 8.5710 - 10 5 _A

log38.46 =L585^_ ^.^^J^'^ ,M

6.9860 - 10 7 .6

1.5850 7^

23 5:^. 24 1^51??.

.2364 ' 5.128

213. Antilogarithms. We can now find the product or quotient
of two numbers if we are able to find the number that corresponds
to a given logarithm.

For this process we have the following

KuLE. If the mantissa is found exactly in the table, the first
two figures of the corresponding number are found in the column
N of the same row, while the third figure of the number is found
at the top of the column in which the mantissa is found.

Place the decimal point so that the rules in § 210 are fulfilled.

EXERCISES

Find the antilogarithms of the following:

1. 3.7419.
Solution: We find the mantissa .7419 in the row which has 55 in coT

umn N. The column in which .7419 is found has 2 at the top. Thus the
significant figures of the antilogarithm are 552. Since the characteristic is 3,
we must by the rule in § 210 have four figures to the left of the point.
Thus the number sought is 5520.

2. 1.3874. 3. 2.7050. 4. .6785.
5. 2.8414.* 6. 5.8831. 7. 1.5752.
8. 9.9112 - 10. ■ 9. 3.7251. 10. 6.3997.

If the mantissa of the given logarithm is between two man-
tissas in the table, we may find the antilogarithm by the following

EuLE. Write the number of three figures corresponding to the
lesser of the two mantissas between which is the given mantissa.

Subtract this mantissa from the given mantissa, and divide
this number by the tabular difference to one decimal place.

Annex this figure to the three already found, and place the
decimal point as the rules in § 210 require.

It should be kept in mind that we may always add and subtract
any integer to a logarithm. This is useful in two cases :

First. When we wish to subtract a larger logarithm from a
smaller ;

Second. When we wish to divide a logarithm by an integer
that is not exactly contained in the characteristic.

Both these processes are illustrated in exercise 2 (1) following.

EXERCISES

1. Find the antilogarithms of the following :

(a) 2.3469.

Solution : The mantissa 3469 is between 3464 and 3483. Hence D = 19.

The mantissa 3464 corresponds to 222. To find the fourth significant figure
of the antilogarithm, divide 3469 - 3464 = 5 by D = 19. Since 5 -^ 19 = .26,
we annex 3 to 222. Hence the antilogarithm = 222.3.

* We write - 2 + .8414 in the form 2.8414 to save space and at the same time to recall
the fact that the mantissa is positive.

LOGARITHMS

245

(b) 4.3147. (c) 1.5271. (d) 1.4216.

(e) 1.6423. (f) 2.8791. (g) .7214.

2. Perform the following operations by logarithms.
1375 X .06423

(a)

76420

Solution :

log 1375= 3.1383
log. 06423= 8.8077- 10

§ 209), 11.9460 - 10
log 76420= 4.8832

Subtracting (Theorem III, § 209), log result = 7.0628 - 10

result = .001156.

(b) (11)8.

(c) iUUY- (d) 5871 -,

- 9308.

(e) (H)«.

(f) (3f I)*-". (g) 7066 -.

-5401.

(h) 8308 X .0003769.

(i) 3410 X .008763.

8.371 X 834.6

,, , 37.42 X 11.21

^■"^ 7309

^^^ BBAl ■

^1) 8/87xV7194
\ 98080000

Solution :

log 87= 1.9395

By Theorem IV, § 209, i log 7194 = 1.9285

= 13.8680 - 10
log 98080000= 7.9916

By Theorem III, § 209, 3)25.8764 - 30

log result = 8.6255 - 10

result = .04222.

Since in the subtraction in this problem we have to subtract 7 from 3,
we add and subtract 10 to the minuend to avoid a negative logarithm. Since
in the division by 3 we would have a remainder in dividing — 10 by 3, we
add and subtract 20 so that 3 may be exactly contained in 30, the negative
part of the logarithm.

(m) ^.

(n) V:06.

(0) ^(.043)8.

(P)^^.

(q) (.21)§.

(r) m V'lOO.

(s) •^:o3.

(t) ^100.

(u) V(l. 563)3.

(v) V.00614.

(w) VHi-

(X) ^0.9 VI!-

(y) ■>J^-v^

(z) >y.47 VM-

214. Cologarithms. The logarithm of the reciprocal of a num-
ber is called its cologarithm. When a computation is to be made

246

in which several numbers occur in the denominator of a fraction,
the subtraction of logarithms is conveniently avoided by the use
of cologarithms. By our definition we have

colog 25 = log 3jV = log 1 - log 25, Theorem III, § 209
log 1 = 10. - 10

log 25 = 1.3979
colog 25= 8.6021-10

Thus in dividing a number by 25 we may subtract the logarithm
of 25, or what amounts to the same thing, add the logarithm of
^ij, which is by definition the cologarithm of 26.

EuLE. The cologarithm of any number is found by subtract-
ing its logarithm from 10 — 10.

In the process of division subtracting the logarithm of a num-
ber and adding its cologarithm are equivalent operations.

EXERCISES

Compute, using cologarithms.
- 8 X 62.73 X .052

66 X 8.793
Solution :

2- MlVlf-
4. V38.462 - 15.382.
g 5086 (.0008769)8
9802 (.001984)* *

</:

' 'XVsox

log 8= .9031

log 62. 73= 1.7975

log. 052= 8.7160-10

colog 56= 8.2518-10

colog 8.793 = 9.0559 - 10
log result = 27.7243 -30

result = .00^99

3. V1572 - 872.

Hint . 1572 - 872 = (157 + 87) (157 - 87)
= (244) (70).

6. V(27:5)2 - (3.483)2.

693 X .04692
03841 X (569.8)2

3 X 421.6 X

.046.(200.1)^

(68.96)^ X 86.61
.09263 V^

1416 X (5.638)2
(75)i

LOGARITHMS 247

215. Change of base. We have seen that the logarithm of a
number for the base 10 may be found to four decimal places in
our tables. It is occasionally necessary to find the logarithm of a
number for a base different from 10. For the sake of generality,
we assume that the logarithms of all numbers for a base b are
computed. We seek a means of finding the logarithm of any
number, as x, for the base c ; that is, we seek to express log^a? in
terms of logarithms for the base h.

Suppose logc X = z, that is, c^ = x.

Take the logarithm of this equation for the base h, and we have

logftC^ = Z log^C = logftOJ.

Then

^ log,a^

l0g6«

If we let

M = l0gf,Gj

we have

.. ^og,x

(1)

This number M does not depend on the particular number x,
but only on the two bases. From (1) we see that we can find the
logarithm of any number for the base c by dividing its logarithm
for the base b by M. The number M is called the modulus of the
new system with respect to the original one.

EuLE. To find the logarithm of a member for a new base c,
divide the common logarithm by the modulus of the system whose
base is c.

EXERCISES

Find:

1. logs 21.

solution: log321.>^-^; = lf^^f = 2.771.
logioo .4771

2. log5 6. 3. logalS.

4. l0gi6 2.

5. logs 167. 6. logi8 237.

7. log2.i6l.41.

248

N.

1

2

3

4

5

6

7

8

9

100

0000

0004

0009

0013

0017

0022

0026

0030

0035

0039

101
102
103

104

105
106

107
108
109

0043
0086
0128

0170
0212
0253

0294
0334
0374

0048
0090
0133

0175
0216
0257

0298
0338
0378

0052
0095
0137

0179
0220
0261

0302
0342
0382

0056
0099
0141

0183
0224
0265

0306
0346
0386

0060
0103
0145

0187
0228
0269

0310
0350
0390

0065

0107.

0149

0191
0233
0273

0314
0354
0394

0069
0111
0154

0195
0237
0278

0318
0358
0398

0073
0116
0158

0199
0241
0282

0322
0362
0402

0077
0120
0162

0204
0245
0286

0326
0366
0406

0082
0124
0166

0208
0249
0290

0330
0370
0410

110

0414

0418

0422

0426

0430

0434

0438

0441

0445

0449

111
112
113

114
115
116

117

118
119

0453
0492
0531

0569
0607
0645

0682
0719
0755

0457
0496
0535

0573
0611
0648

0686
0722
0759

0461
0500
0538

0577
0615
0652

0689
0726
0763

0465
0504
0542

0580
0618
0656

0693
0730
0766

0469
0508
0546

0584
0622
0660

0697
0734
0770

0473
0512
0550

0588
0626
0663

0700
0737
0774

0477
0515
0554

0592
0630
0667

0704
0741
0777

0481
0519
0558

0596
0633
0671

0708
0745
0781

0484
0523
0561

0599
0637
0674

0711
0748

0785

0488
0527
0565

0603
0641
0678

0715
0752
0788

120

0792

0795

0799

0803

0806

0810

0813

0817

0821

0824

121
122
123

124

125
126

127

128
129

0828
0864
0899

0934
0969
1004

1038
1072
1106

0831
0867
0903

0938
0973
1007

1041
1075
1109

0835
0871
0906

0941
0976
1011

1045
1079
1113

0839
0874
0910

0945
0980
1014

1048
1082
1116

0842
0878
0913

0948
0983
1017

1052
1086
1119

0846
0881
0917

0952
0986
1021

1055
1089
1123

0849
0885
0920

0955
0990
1024

1059
1093
1126

0853
0888
0924

0959
0993
1028

1062
1096
1129

0856
0892
0927

0962
0997
1031

1065
1099
1133

0860
0896
0931

0966
1000
1035

1069
1103
1136

130

1139

1143

1146

1149

1153

1156

1159

1163

1166

1169

131
132
133

134
135
136

137

138
139

1173
1206
1239

1271
1303
1335

1367
1399
1430

1176
1209
1242

1274
1307
1339

1370
1402
1433

1179
1212
1245

1278
1310
1342

1374
1405
1436

1183
1216
1248

1281
1313
1345

1377
1408
1440

1186
1219
1252

1284
1316
1348

1380
1411

1443

1189
1222
1255

1287
1319
1351

1383
1414
1446

1193
1225
1258

1290
1323
1355

1386
1418
1449

1196
1229
1261

1294
1326
1358

1389
1421
1452

1199
1232
1265

1297
1329
1361

1392
1424
1455

1202
1235
1268

1300
1332
1364

1396
1427
1458

140

1461

1464

1467

1471

1474

1477

1480

1483

1486

1489

141
142
143

144
145
146

147
148
149

1492
1523
1553

1584
1614
1644

1673
1703
1732

1495
1526
1556

1587
1617
1647

1676
1706
1735

1498
1529
1559

1590
1620
1649

. 1679
1708
1738

1501
1532
1562

1593
1623
1652

1682
1711
1741

1504
1535
1565

1596
1626
1655

1685
1714
1744

1508
1538
1569

1599
1629
1658

1688
1717
1746

1511
1541
1572

1602
1632
1661

1691
1720
1749

1514
1544
1575

1605
1635
1664

1694
1723
1752

1517
1547

1578

1608
1638
1667

1697
1726
1755

1520
1550
1581

1611
1641
1670

1700
1729
1758

150

1761

1764

1767

1770

1772

1775

1778

1781

1784

1787

N.

1

2

3

4

5

6

7

8

9

LOGARITHMS

249

N.

1

2

3

4

5

6

7

8

9

150

1761

1764

1767

1770

1798
1827
1855

1884
1912
1940

1967
1995
2022

1772

1775

1778

1781

1784

1787

151
152
153

154
155
156

157

158
159

1790
1818
1847

1875
1903
1931

1959
1987
2014

1793
1821
1850

1878
1906
1934

1962
1989
2017

1796
1824
1853

1881
1909
1937

1965
1992
2019

1801
1830
1858

1886
1915
1942

1970
1998
2025

1804
1833
1861

1889
1917
1945

1973
2000
2028

1807
1836
1864

1892
1920
1948

1976
2003
2030

1810
1838
1867

1895
1923
1951

1978
2006
2033

1813
1841
1870

1898
1926
1953

1981
2009
2036

1816
1844
1872

1901
1928
1956

1984
2011
2038

160

2041

2044

2047
2074
2101
2127

2154
2180
2206

2232

2258
2284

2049

2052

2055

2057

2060

2063

2066

161
162
163

164
165
166

167
168
160

2068
2095
2122

2148
2175
2201

2227
2253
2279

2071
2098
2125

2151
2177
2204

2230
2256
2281

2076
2103
2130

2156
2183
2209

2235
2261

2287

2079
2106
2133

2159
2185
2212

2238
2263
2289

2082
2109
2135

2162
2188
2214

2240
2266
2292

2084
2111
2138

2164
2191
2217

2243
2269
2294

2087
2114
2140

2167
2193
2219

2245
2271

2297

2090
2117
2143

2170
2196
2222

2248
2274
2299

2092
2119
2146

2172
2198
2225

2251
2276
2302

170

2304

2307

2310

2312

2315

2317
2343
2368
2393

2418
2443
2467

2492
2516
2541

2320

2322

2325

2327

171
172
173

174
175
176

177

178
179

2330
2355
2380

2405
2430
2455

2480
2504
2529

2333
2358
2383

2408
2433
2458

2482
2507
2551

2335
2360
2385

2410
2435
2460

2485
2509
2533

2338
2363
2388

2413
2438
2463

2487
2512
2536

2340
2365
2390

2415
2440
2465

2490
2514

2538

2345
2370
2395

2420
2445
2470

2494
2519
2543

2348
2373
2398

2423
2448
2472

2497
2521
2545

2350
2375
2400

2425
2450
2475

2499
2524
2548

2353
2378
2403

2428
2453
2477

2502
2526
2550

180

2553

2555

2558

2560

2562

2565

2567

2570

2572

2574

181
182
183

184

185
186

187
188
189

2577
2601
2625

2648
2672
2695

2718
2742
2765

2579
2603
2627

2651
2674
2697

2721
2744
2767

2582
2605
2629

2653
2676
2700

2723
2746
2769

2584
2608
2632

2655
2679
2702

2725

2749

2772

2586
2610
2634

2658
2681
2704

2728
2751
2774

2589
2613
2636

2660
2683
2707

2730
2753
2776

2591
2615
2639

2662
2686
2709

2732
2755

2778

2594
2617
2641

2665
2688
2711

2735

2758
2781

2596
2620
2643

2667
2690
2714

2737
2760
2783

2598
2622
2646

2669
2693
2716

2739
2762
2785

190

2788

2790

2792

2794
2817
2840
2862

2885
2907
2929

2951
2973
2995

2797

2799

2801

2804

2806

2808

191
192
193

194
195
196

197
198
199

2810
2833
2856

2878
2900
2923

294.:
2967
2989

2813
2835
2858

2880
2903
2925

2947
2969
2991

2815
2838
2860

2882
2905
2927

2949
2971
2993

2819
2842
2865

2887
2909
2931

2953
2975
2997

2822
2844
2867

2889
2911
2934

2956
2978
2999

2824
2847
2869

2891
2914
2936

2958
2980
3002

2826
2849
2871

2894
2916
2938

2960
2982
3004

2828
2851
2874

2896
2918
2940

2962
2984
3006

2831
2853
2876

2898
2920
2942

2964
2986
3008

200

3010

3012

3015

3017

3019

3021

3023

3025

3028

3030

N.

1

2

3

4

5

6

7

8

9

250

N.

1

2 3

4

5

6

7

8

M

20

3010

3032

3054

3075

3096

3118

3139

3160

3181

320r

21

22
23

24
25
26

27
28
29

3222
3424
3617

3802
3979
4150

4314
4472
4624

3243
3444
3636

3820
3997
4166

4330
4487
4639

3263
3464
3655

3838
4014
4183

4346
4502
4654

3284
3483
3674

3856
4031
4200

4362
4518
4669

3304
3502
3692

3874
4048
4216

4378
4533
4683

3324
3522
3711

3892
40(35
4232

4393
4548
4698

3345
3541
3729

3909
4082
4249

4409
4564
4713

3365
3560
3747

3927
4099
4265

4425
4579

4728

3385
3579
3766

3945
4116
4281

4440
4594
4742

3404
3598
3784-

3962
4133
4298

4456
4609
4757

30

4771

4786

4800

4814

4829

4843

4857

4871

4886

4900

31
32
33

34
35
36

37
38
39

4914
5051
5185

5315
5441
5563

5682
5798
5911

4928
5065
5198

5328
5453
5575

5694
5809
5922

4942
5079
5211

5340
5465
5587

5705
5821
5933

4955
5092
5224

5353
5478
5599

6717
5832
5944

4969
5105
5237

5366
5490
5611

5729
5843
5955

4983
5119
5250

5378
5502
5623

5740

6855
5966

4997
5132
6263

5391
5514
5635

5752
5866
5977

6011
6146
5276

5403
5527

6647

6763

5877
6988

5024
5159
5289

5416
6639
6658

5775
5888
5999

5038
5172
5302

5428
5551
5670

6786
6899
6010

40

6021

6031

6042

6053

6064

6075

6085

6096

6107

6117

41
42
43

44
45
46

47

48
49

6128
6232
6335

6435
6532
6628

6721
6812
6902

6138
6243
6345

6444
6542
6637

6730
6821
6911

6149
6253
6355

6454
6551
6646

6739
6830
6920

6160
6263
6365

6464
6561
6656

6749
6839
6928

6170
6274
6375

6474
6571
6665

6758
6848
6937

6180
6284
6385

6484
6580
6675

6767
6857
6946

6191
6294
6395

6493
6590
6684

6776
6866
6955

6201
6304
6405

6503
6599
6693

6786
6875
6964

6212
6314
6415

6513
6609
6702

6794
6884
6972

6222
6325
6425

6522
6618
6712

6803
6893
6981

50

6990

6998

7007

7016

7024

7033

7042

7050

7059

7067

51
52
53

54
55
56

57

58
59

7076
7160
7243

7324
7404

7482

7559
7634
7709

7084
7168
7251

7332
7412
7490

7566
7642
7716

7093
7177
7259

7340
7419
7497

7574
7649
7723

7101
7185
7267

7348
7427
7505

7582
7657
7731

7110
7193
7275

7356
7435
7513

7589
7664
7738

7118
7202
7284

7364
7443

7620

7597
7672
7745

7126
7210
7292

7372
7451

7528

7604
7679
7752

7135
7218
7300

7380
7469
7636

7612
7686
7760

7143
7226
7308

7388
7466
7543

7619
7694
7767

7152
7235
7316

7396

7474
7551

7627
7701
7774

60

7782

7789

7796

7803

7810

7818

7826

7832

7839

7846

61
62
63

64
65
66

67
68
69

7853
7924
7993

8062
8129
8195

8261
8325
8388

7860
7931
8000

8069
8136
8202

8267
8331
8395

7868
7938
8007

8075
8142
8209

8274
8338
8401

7875
7945
8014

8082
8149
8215

8280
8344
8407

7882
7952
8021

8089
8156

8222

8287
8351
8414

7889
7959
8028

8096
8162
8228

8293
8367
8420

7896
7966
8036

8102
8169
8235

8299
8363
8426

7903
7973
8041

8109
8176
8241

8306
8370
8432

7910
7980
8048

8116
8182
8248

8312
8376
8439

7917
7987
8055

8122

8189
8254

8319
8382
8445

70

8451

8457

8463

8470

8476

8482

8488

8494

8500

8506

N.

1

2

3

4

5

6

7

8

9

LOGARITHMS

251

N.

1

2

3

4

5

6

7

8

9

70

8451

8457

8463

8470

8476

8482

8488

8494

8500

8506

71
72
73

8513
8573
8633

8519
8579
8639

8525

8585
8645

8531
8591
8651

8537
8597

8657

8543
8603
8663

8549
8609
8669

8555
8615
8675

8561
8621
8681

8667
8627
8686

74

75
76

8692
8751
8808

8698
8756
8814

8704
8762
8820

8710
8768
8825

8716

8774
8831

8722
8779
8837

8727
8785
8842

8733
8791
8848

8739
8797
8854

8745
8802
8859

77
78
79

8865
8921
8976

8871
8927
8982

8876
8932
8987

8882
8938
8993

8887
8943
8998

8893
8949
9004

8899
8954
9009

8904
8960
9015

8910
8965
9020

8915
8971
9025

80

9031

9036

9042

9047

9053

9058

9063

9069

9074

9079

81
82
83

9085
9138
9191

9090
9143
9196

9096
9149
9201

9101
9154
9206

9106
9159
9212

9112
9165
9217

9117
9170
9222

9122
9175
9227

9128
9180
9232

9133
9186
9238

84
85
86

9243
9294
9345

9248
9299
9350

9253
9304
9355

9258
9309
9360

9263
9315
9365

9269
9320
9370

9274
9325
9375

9279
9330
9380

9284
9335
9385

9289
9340
9390

87
88
89

9395
9445
9494

9400
9450
9499

9405
9455
9504

9410
9460
9509

9415
9465
9513

9420
9469
9518

9425
9474
9523

9430
9479
9528

9435
9484
9533

9440
9489
9538

90

9542

9547

9552

9557

9562

9566

9571

9576

9581

9586

91
92
93

9590
9638
9685

9595
9643
9689

9600
9647
9694

9605
9652
9699

9609
9657
9703

9614
9661
9708

9619
9666
9713

9624
9671
9717

9628
9675
9722

9633
9680
9727

94
95
96

9731
9777
9823

9736

9782
9827

9741
9786
9832

9745
9791
9836

9750
9795
9841

9754
9800
9845

9759
9805
9850

9763
9809
9854

9768
9814
9859

9773
9818
9863

97

98
99

9868
9912
9956

9872
9917
9961

9877
9921
9965

9881
9926
9969

9886
9930
9974

9890
9934
9978

9894
9939
9983

9899
9943
9987

9903
9948
9991

9908
9952
9996

100

0000

0004

0009

0013

0017

0022

0026

0030

0035

0039

N.

1

2

3

4

5

6

7

8

9

216. Exponential equations. Equations in which the variable
occurs only in the exponents may often be solved by the use of
tables of logarithms if one keeps in mind the fact that
log oc^ = ic log a.

EXERCISES

Solve the following :

1. 10^-1 = 4.

Solution : Taking the logarithm of both sides of the equation, we have
(x-1) log 10 = log 4,
or since log 10 = 1, a; = log 4 + 1 = .G021 + 1 = 1.0021.

2. 4* -3^ = 8,
2* . 81/ = 9.

Solution : Taking the logarithms of the equations, we have

a;log4 + ylogS = log 8,
a: log 2 + y log 8 = log 9,

or jc21og2 4-ylog3 = 31og2, (1)

a;log2 + y31og2 = 21og3.
Eliminate x.

x21og2+ 2/log3 = 31og2

a;21og2 + y61og2 = 41og3

2/(log3 - 61og2) = 31og2 - 41og3

_ 31og2-41og3 _ 3 X .3010-4 x .4771
^~ Iog3-01og2 ~ .4771 - 6 X .3010

.9030 - 1.9084 _ - 1.0054 _ 1.0054
~ .4771 - 1.8060 ~ - 1.3289 ~ 1.3289*

Perform this division by logarithms.

log 1.0054 = 10.0023 - 10
log 1.3289= .1235

logy= 9.8788-10
y= .7565.
Substituting in (1),

a; = 31og2 - .7565 log3 _ .9030 - .7565 x .4771
2 log 2 ~ .6020

Compute .7666 x .4771 by logarithms.

log. 7565= 9.8788-10
log. 4771= 9.6786-10 .
log result = 19. 5574 - 20
result = .3609.

„ .9030 -.3609 .5421

Hence x — =

.6020 .6020

log.5421 = 19.7341 -20

log. 6020= 9.7796 - 10

log;c= 9.9545- 10

X = .9005.

3. 6* = 2. 4. 4y = 3. 5. 7*+« = 5.

6. 32x + i_5. 7.4^-1 = 6^+1. 8. 22a:+3_ 6^-1 = 0.

a^.6y = m, -^ a^.6!' = w,

c* . d?/ = 71. ' x-\- y — n.

LOGARITHMS 253

-- 2^. 2^ = 222, -2 a2^-3.a3y-2:=a8.

' x-y = A. 8x + 2y = n.

- 3 3* • 4y = 15562, .. ^^2^^ . ^/^^F^i = a\

4^ . 52/ = 128000. 4,^:,— ^ 3

^l)3x + 5 - ^sjb^y + 1 = 610.

217. Compound interest. If |1500 is at the yearly interest
of 3%, the total interest for a year is \$1500 • (0.03) = \$45. The
total sum invested at the end of a year would be \$1545.

Let, in general, P represent a sum of money in dollars.

Let r represent a yearly rate of interest.

Then P • r represents the yearly interest on P, and
P-f p.r = P(r + 1)
represents the total investment, principal and interest, at the
end of a year.

Similarly, P (r + 1) r is the second year's interest, and

P{r + l)r + P(r + 1) = P(r2 + r + r + 1) = ^(^ + 1)^
is the total investment at the end of two years.

In general, A = P (r + 1)" (1)

is the total accumulation at the end of tz years. If we know r, P,
and n, we can by (1) find A. If we take the logarithm of both
sides of the equation, we have

log A = log P -\- n log (r + 1),

log^-logP ^ (2)

log(r + l) ^ ^

Hence if we know A, P, and r, we can find n.
If the interest is computed semiannually, we have as interest

v

at the end of a half-year P • ^ ' while the entire sum would be

P (- 4- 1 ) . Reasoning as above, we find that if the interest is com-
puted semiannually, the accumulation at the end of n years is

^=pg + l)" (3)

Similarly, n= l"g -^ ' l»g f . (4)

logg + l)

3

If the interest is computed k times a year, we liave at the end
of n years

A-P(^ + 1) , (5)

Hogg + l)

EXERCISES

In such exercises as the following, four-place tables are not sufficiently
exact to obtain perfect accuracy. In general, the longer the term of years
and the more frequent the compounding of interest, the greater the inaccuracy.

1. If \$1600 is placed at 3|% interest computed semiannually for 13 years,
to how much will it amount in that time ?

Hence

ion: By formula (3),

A =

4

-)

2n

P = 1600, r =

.03|,

n =

13

36 A = 1600 {-^

\26

1600

\400

-)^=-Q^-

log 1600= 3.2041

26 log407 = 67.8496

71.0537

261og400 = 67.6546

log^= 3.3991

A = \$2507.

Iog407 = 2.e096

26

156576

52192

67.8496

log 400 =2.6021

26

156126

52042

67.6546

2. After how long will \$600 at 6% computed annually amount to \$1000 ?

Solution : By formula (2) we have

_ log J. — logP

log(r + l)

A = 1000, P = 600, r = .06.

log 1000 - log 600 3 - 2.7782 .2218 _ ^,.

n = -^ = =^ = 8./0 years.

log 1.06 .0253 .0253 ^

.76 year = .76 • 12 = 9.12 months.

.12 month = .12 • 30 = 3.6 days.

Thus n = 8 years 9 months 3.6 days.

LOGARITHMS 255

In the following exercises the interest is computed annually unless the
contrary is stated.

3. To what will \$3750 amount in 20 years if left at 6% interest ?

4. To what sum will \$25,300 amount in 10 years if left at 5% interest
computed semiannually?

5. To what does \$1000 amount in 10 years if left at 6% interest computed
(1) annually, (2) semiannually, (3) quarterly ?

6. A sum of money is left 22 years at 4% and amounts to \$17,000. How
much was originally put at interest ?

7. What sum of money left at 4|% for 30 years amounts to \$30,000 ?

8. What sum of money left 10 years at 4i% amounts to the same sum as
\$8549 left 7 years at 5% ?

9. If a man left a certain sum 11 years at 4%, it would amount to \$97 less
than if he had left the same sum 9 years at 5%. What was the sum ?

10. Which yields more, a sum left 10 years at 4% or 4 years at 10%?
What is the difference for \$1000 ?

11. Two sums of money, \$25,795 in all, are left 20 years at 4f%. The
difference in the sums to which they amount is \$14,660. What were the sums ?

12. At what per cent interest must \$15,000 be left in order to amount to
\$60,000 in 32 years ?

13. At what per cent must \$3333 be left so that in 24 years it will
amount to \$10,000 ?

14. Two sums of which the second is double the first but is left at 2% less
interest amount in 36 1 years to equal sums. At what per cent interest was
each left ?

15. In how many years will a sum double if left at 5% interest ?

16. In how many years will a sum double if left at 6% interest computed
semiannually ?

17. In how many years will a sum amount to ten times itself if left at
4% interest ?

18. In how many years will \$17,000 left at 4^% interest amount to the
same as \$7000 left at 5|% for 20 years ?

19. On July 1, 1850, the sum of \$1000 was left at 4i% interest. When
paid back it amounted to \$2222. When did this occur ?

20. Prove formulas (1), (3), and (5) by complete induction.

CHAPTEE XXI
CONTINUED FRACTIONS

218. Definitions. A fraction in the form

e-¥f

where a, &,•••, ^, •• • are real numbers, is called a continued fraction.
We shall consider only those continued fractions in which tlie
numerators b, d, /, etc., are equal to unity and in which the
letters represent integers, as for example :

»! H . written a^-\ — —

When the number of quotients ^2? ^s? ^4? • • • is finite the frac-
tion is said to be terminating. When the fraction is not terminat-
ing it is infinite. We shall see that the character of the numbers
represented by terminating fractions differs widely from that of
the numbers represented by infinite continued fractions. We shall
find, in fact, that any root of a linear equation in one variable,
i.e. any rational number, may be represented by a terminating
continued fraction, and conversely; furthermore, that any real
irrational root of a quadratic equation may be represented by the
simplest type of infinite continued fractions, and conversely.

219. Terminating continued fractions. If we have a terminat-
ing continued fraction, where ^i, a2j • • • are integers, it is evident
that by reducing to its simplest form we obtain a rational num-
ber. The converse is also true, as we can prove in the following

256

CONTINUED FRACTIONS 257

Theorem. Any rational number may he expressed as a ter-
minating fraction.

Let - represent a rational number. Divide a by J, and let a^

be the quotient and c (which must be less than h) the remainder.
Then (§ 26) ,

-^a, + - = a, + j

Divide h by c, letting ^g t)® the quotient and d (which must be
less than c) the remainder. Then

- = «! H

b a^ -\- d

c

Continuing this process, the maximum limit of the remainders
in the successive divisions becomes smaller as we go on, until
finally the remainder is zero. Hence the fraction is terminating.
It is noted that the successive quotients are the denominators in
the continued fraction.

EXERCISES

1. Convert the following into continued fractions

(a) ^VV-

Solution: 247|77[0

77J247|3
231
16J77[4
64

13J16[1
13

3J13[4
12
1J313
3

The continued fraction is

ZL-1 1 1 1 1.

247 ~ 3 + 4 + 1 + 4 + 3*

(b) if. (c) ^Vt. (d) m-

(e) iM- (f) Iff. (g) /jVf-

(h) HH- (i) Hf. (J) mh

2. Express the following continued fractions as rational fractious.

^''^2 + 3 ^"^l + r ^"^-a + l-

(d)l I. (e)i 11. (f)l 1 1.

^ ' a; + x ^ M+2 + 3 ^ ' 3 + 4 + 5

,.11111 ,^111111

(g) ----- . (h) ------ .

^^2+4+2+4+2 ^^1+2+3+1+2+3

220. Convergents. The value obtained by taking only the first
n ~ 1 quotients in a continued fraction is called the nth. convergent
of the fraction.

Thus iu the fraction

1+1111

2+3+2+6 :

1 is the first convergent,

1 3

1 + - = - is the second convergent,

■a 2t

l + -=l+- = — is the third convergent, etc.

3
When there is no whole number preceding the fractional part of the continued
fraction the first convergent is zero. Thus in

111

2 + 3 + 5
\ is called the second convergent.

In the continued fraction

,1111

^2 + ^3 + ^4 + «5 H

let -i, -1, -i, . .. represent the successive convergents expressed

9.\ ?2 ^z
as rational fractions.

Then for the first convergent we have

^ = — > or^i = ai, g'i = l.
9.\

CONTINUED FRACTIONS 259

For the second convergent we have

, 1 a^a^ + 1 J92 , H , ^

«! H = = — > or ^2 = «^2<^l + 1 = «2i?i + 1,

^2 <^2 2'2

!Z2 = «2 = ^22'l-

For the third convergent we have

^2 + 1^ a3«^2 + 1 «^3«2 +1 2'3

«3

or i?3 = % (^2% + 1) 4- ^1 = ^3^2 +i?l,

?Z3 = «3«2 + 1 = %'Z2 + (1\-

This indicates that the form of the r^th convergent is

9'n a«2'n-l + 2'»-2 ^

This is in fact the case, as we proceed to show by complete
induction.

We have already established form (1) for n = 2 and n = ^.
We assume it for n = m, and will show that its validity for
n = m -\-l follows. The (m + l)th convergent differs from the

mth only in the fact that a^ -\ appears in the continued

fraction in place of a^. In (1) replace n by m, and a„ by

a^ H J and we have

Pr

1+1 __ \ *^m+l/

'■^^ Um + ) 2'm-l + 2'm-2

\ «'m+l/

_ (Q^m-H<^>n + ^)Pm^l + Q^„. + 1 J^m - 2
(am+;iam + l).?m-l + «m + l?m-2

_ Q^m4-lKi>m-l+i>m~2)+i>»»-l
«^». + l(am2'm-l + 9'm-2)+ 2'm-l

which is form (1).

EXERCISES

1. Express the following as continued fractions, and find the convergents.

(a) !?.

Solution : By the method already explained, we find that

30 _1 1 1 1 1 1

il~l + 2+l + 2 + l + 2'
Here ai = 0, a2 = 1, as = 2, 04 = 1, as = 2, aQ = 1, a? = 2.
The first convergent is evidently 0, the second is 1, and the third is
1 2

mi- ir ^. ^ - P4 CliPS +P2 1-2+1 3

The fourth convergent is — = -^^ — — = = -.

Qi a^qs + ga 1-3 + 1 4

mu ^x^u . • P5 ttsP* + P3 2-3 + 2 8

The fifth convergent is — = —^ — ^ = = — •

?5 a^qi + ^3 2-4 + 3 11

aePn + P4 1-8 + 3 11

The sixth convergent is ^^ =
The seventh convergent is

^6 tteO's + 5'4 1-11 + 4 15
Pi OtPg+Ps 2-11 + 8 30

(b) 3¥2. (c) If. (d) t¥t- (e) tVt- (f) t?7-

Qi a^q& + q^ 2-15 + 11 41

(f) \'-'-

(g)TVj. (b)TW^. (i)?\V U) mV

2. Find the value of the following by finding the successive convergents.

,,11111 ,^,11111

(a) - - - - -• (b) - - - - -.

^^2+1+2+1+2 ^'3+2+3+2+3

(c^ 1 i i i i i (d^ 1 1 1 1 1 1

^^^2 + 3+1 + 1 + 3 + 2* ^^3 + 3_^34.34.34.3'

(e^ 1 1 ' 1 1 1 1 (f\ 1 1 1 1 1 1

^^^6 + 3 + 1 + 1 + 3 + 6' ^^1 + 34.5+5 + 3+1'

1 1 1 1 1 1

^^^ (X - 1) + X + (X + 1) * ^ ^ X + X + x'

221. Recurring continued fractions. We have seen that e very-
terminating continued fraction represents a rational number, and
conversely. We now discuss the character of the numbers repre-
sented by the simplest infinite continued fractions. A recurring
continued fraction is one in which from a certain point on, a
group of denominators is repeated in the same order.

CONTINUED FRACTIONS 261

----- ^

^^ 3 + 2 + 3 + 2 + 3+2 + *"'

111111

1+2+3+1+2+3+'"

are recurring continued fractions if the denominators are assumed to repeat
indefinitely as indicated.

That a repeating continued fraction actually represents a num-
ber we shall establish in § 223. Unless this fact is proven, one
runs the risk of dealing with symbols which have no meaning.
If for certain continued fractions the successive convergents
increase without limit, or take on erratic values that approach no
limit, it is important to discover the fact. All the fractions that
we discuss actually represent numbers, as we shall see.

We shall consider only continued fractions in which every
denominator has a positive sign.

Theorem. Every recurring continued fraction is the root of a

T w • . 111111

Let, tor instance, a; = - - - - - - •••.

' ' a -\- b -\- c -\- a -\- b -\- G -[-

Evidently the part of the fraction after the first denominator c
may be represented by cc, and we have thus virtually the termi-
nating fraction

X

a -\- b -\- c -\- X

The second convergent is —
The third convergent is

ab -\-l q^
The fourth convergent, or a?, gives us

^jP4^ CT4^3+-^2 ^ (c+-a;)5+-l
q^ ci^qz 4- S'2 (c + x) (ab +- 1) 4- fit
Simplifying, we get

(ab + l)x^-{- lc(ab -{-l)-{- a - b'jx - be -1 = 0,

which is a quadratic equation whose root is a?, the value of the
continued fraction.

Since this equation has a negative number for its constant term
it has one positive and one negative root. The continued fraction
must represent the positive root, since we assume that the letters
la, h, c represent positive integers. The quadratic equation whose
root is a recurring continued fraction with positive denomina-
tors will always have one positive and one negative root. The
equation will be quadratic, however, whatever the signs of the
denominators may be.

The proof may be extended to the case where there are any
number of recurring denominators or any number of denominators
before the recurrence sets in. Since every real irrational root
of a quadratic equation is a surd, our result is equivalent to the
statement that every recurring continued fraction may be ex jft*essed
as a surd.

EXERCISES

Of what quadratic equations are the following roots ? Express the con-
tinued fraction as a surd.

ill 1 1

* 2 + 3 + 2 + 3 + ""'

Solution : Let x = - - .

2 + 3 + a:

Then x = l 1 ^ + ^

2+- 6+2X+1

3 + «

or 2 0:2 + 6 X - 3 = 0.

Solving this equation, we get

- 3 + VT6 - 3 - Vl5
X\ = or X2 =

2 2

Since x^ is negative, xi must be the surd that is represented by the con-
tinued fraction.

1111 _3 + Vl6

Thus
2.

2+3+2+3+ 2

1111 3IIII

1 + 2 + 1 + 2 + "*' ■ 3 + 2 + 3 + 2 +

6.2 + 1 1 1 1 .

2+1+2+1+

•••

m.r.u. = 2 + l^\^.

••.

then ^-^=l^\^--

and . = 2 + i^-;^(^.

-2)

CONTINUED FRACTIONS 263

^111111 silil

• 1 + 2 + 3 + 1+2 + 3 + "" •3+1 + 3 + 1 + *"'

« o 1 1 1

7. 3 + - - - ....

^3+4+5+

8.1 + 1 1 1 1

3+4+3+4+

9. 1 + i 1 1 i i 1 ....

1+2+3+1+2+3+

10. i 1 1 -1 1 i ....

2+1+2+2+1+2+

222. Expression of a surd as a recurring continued fraction.

This is the converse of the problem discussed in the last sec-
tion, and shows that recurring continued fractions and quadratic
equations are related in the same intimate way that terminat-
ing fractions and rational numbers (i.e. the roots of linear equa-
tions) are connected. We seek to express an irrational number,
as, for instance, V2, as a continued fraction. This we may do
as follows.

Since 1 is the largest integer in V2 we may write

V2 = l + (V2-l) = l+(-^^^^.
Rationalizing the numerator, we have

V2 = i + --i —

V2+-1

Since 2 is the largest integer in V2 +- 1 we have

V2 = l+ 7^= 7 = 1 +

2+-(V2-l) o I (V2-I)

Rationalizing the numerator V2 — 1, we have

^"^viTI ^ + 2+(V2-i)

By continuing this process we continually get tlie denomi-
nator 2. Thus 11

This process consists of the successive application of two opera-
tions, and affords the

EuLE. Express the surd as the sum of two numbers the first
of which is the largest integer that it contains.

nationalize the numerator of the fraction whose numerator is
the second of these numbers. Repeat these operations until a
recurrence of denominators is observed.

This process may be applied to any surd, and a continued frac-
tion which is recurring will always be obtained. We shall con-
tent ourselves with a statement of this fact without proof.

If the surd is of the form a — V^, a continued fraction may be
derived for + V^ and its sign changed. Since the real roots of
any quadratic equation cc^ + 2 a^x + ag = are surds of the form
a ± V^, where a and b are integers, it appears that the roots of
any such equation may be expressed as recurring continued frac-
tions. It can be shown that the real roots of the general quad-
ratic equation a-o^^ + ajcc -f ctg = may also be so expressed.

EXERCISES

1. Express the following surds as recurrent continued fractions,
(a) 2 + Vs.
Solution :

2+V3 = 3 + (V3-l) = 3+ (1)

= 34^^ = 3+ 2'

V3+ 1 V3+I

= 3 + ^^^ — = 3 + J = 3 +

V3 + I ^ V3+I _ . V3-I

2 .2 "^2

3 + 1^ 3-1 =3 + ^^^ = 3 + 1 1

2(V3 + 1) V3 + I 2-|-(V3-l)

1

conti:nued fkactions 265

But since Vs — 1 is the same number that we have in (1), this fraction
repeats from this point on, and we have

2+V3 = 8 + l 1 I 1 ....
1+2+1+2+

(b) V5.

(c) Vl7.

(d) V65.

(e) V47.

(f) Vli.

(g) V23.

(h) V3i.

(i) Vl9.

(J) V62.

(k) V79.

(1) V98.

(m) V88.

(n) V22.

(0) Vis.

(p) V59.

(q) VlOl.

(r) 7 + VII.

(s) 8-

-V3

(t)

3-V23.

2. Express as a continued fraction the roots of the following equations,
y (a) a;2 - 7 X - 3 = 0. / (b) x^ + 2 a; - 6 = 0.

(c) x2 + 3x - 8 = 0. (d) x2 - 4x - 4 = 0.

223. Properties of conver gents. The law of formation of con-
vergents given in § 220 is valid whether the continued fraction
is terminating or infinite. We should expect that in the case of
an infinite fraction the successive convergents would give us an
increasingly close approximation to the value of the fraction.
This is indeed the fact, as we shall see.

Theorem. The difference between the nth and (n + l)st con-
vergents is ^

Mn + l

We prove this theorem by complete induction.
Let the continued fraction be

a, H — —

«2 + ^3 + »4 +

Then the first and second convergents are respectively
Then £2-a=(^i^L±l)_„, = i.

2'2 S'l »2 «2

(1)

Since ^i = 1? 2'2 = ^2?

we have ^zl±i _^ = izL^yHl for n = 1.

We assume that the theorem holds for n — m^ that is,

Prn±l _ ^ -y^m^m + i+gmPm + l ^ {zlJ}!^.

We must prove that it holds for n = m -{• 1,

Now since - ^"^^^ — - ^'" + ^ ^ Pm + l^m + 2 YPm + 29^m+l^
2'm + 2 S'm + l 2'm + l2'»i + 2

our theorem reduces to proving that the numerator

-- Pm+lQm + 2 ■^Pm + 2^m + l=(- l)""^"- (2)

In the left-hand member of (2) set

«m + 2'7m+l + ^m = 2'm + 2, (1)> § 220

and «^m + 2i^m + l+i^m=I>m + 2-

Then -- Pm+li^m+^^m + l H- 2'm)f (^m+2P„.+ l +i?».)?n. + l
=^m+l^m-^Pm<Im+l = - (Pm'Ln+ I - Pm + l^m)

Corollary I. The difference between the successive convergents
of a continued fraction with positive denominators approaches
zero as a limit.

Since q^ = a^q„_-^ + q^-^, evidently q^ increases without limit
when n is increased, since to obtain a^ we add together positive
numbers neither one of which can vanish.

Thus we can find a value of n large enough so that — j and

1 . ^« .

hence ? will be smaller than any assigned number, which

is another way of stating that as n increases approaches

zero as a limit. ^'•^'* "*" ^

CONTINUED FliACTIONS 267

Corollary II. The even convergents decrease, while the odd
convergents increase, as n increases.

We must show that

PTn + 2 Pm

9.m + 2 5'm

is negative or positive according as ni is even or odd. Adding
and subtracting ? we have

?/m + 1

Pm + 2 Pm (Pjn±2 Pin±\ i ,

m+l

2'm + 2 ^m Vlm + 2 <!

/Pjn±l_ln\
X^m + l ^mj

_ (-1)"»+^ (-1)"* + ^

By Corollary I, the denominator of the first fraction exceeds
that of the second. Hence when m is odd the sum in the last
member of the equation is positive, and when 7n is even the sum
is negative.

We now se* that any recurring fraction of the type considered
in § 221 actually represents a number in the sense of § 74. We
have seen that the successive odd convergents continually increase,
while the even convergents continually decrease, until the differ-
ence between a pair of them is very small. Such sequences of
numbers we have seen (§73 ff.) define real numbers.

224. Limit of error. We are now in a position to state a maxi-
mum value for the error made in taking any convergent of a con-
tinued fraction for the fraction itself.

Theorem. The maximum limit of error in taking the nth
convergent for the continued fraction is less than

Since by the theorem of the last section the value of the frac-
tion is between any pair of consecutive convergents, it must
differ from either of these convergents by less than they differ

from each other, that is, by less than

EXERCISES
Find a convergent that differs by less than .001 from each of the following ;

1. V6.
Solution :

V6 = 2 + ( V6 - 2) = 2 + ^_~^ = 2 +

V6 + 2 V6'+2

2

= 2 + i 6-4 =2 + 5 1 =^ + lj_l

Since the last surd repeats the one in the first equation we have

V6 = 2 + i 1 1 1 ....

2+4+2+4+

Since

Pi _ 2 . J92 _ 5 . Pa _ 4 • 5 + 2 _ 22 .
gi ~ 1 ' 52 ~ 2 ' ^3 4 . 2 + 1 ~ 9 '

P4 2-22 + 5 49. ps 4-49 + 22
g4~2.9 + 2~20' 95~4-20 + 9~

218
89

1-^-1 <.001.

54^5 20 - 89 1780

we see by § 224 that |f satisfies the condition, of the problem.

2. V7. 3. V46. , 4. Vs. 5. Vl9.

6. V36. 7. V32. 8. V6l. 9. V65.

10. S+V2.- 11. V99. 12. Vil. 13. Vl3.

14. The number ir has the value 3. 14169. Find by the method of continued
fractions a series of convergents the last of which differs from this value by
less than .OCioi.

CHAPTEE XXII

INEQUALITIES

225. General theorems. We say that a is greater than h when
a — 6 is positive. If a — h is negative, then a is less than h.
Thus any positive number or zero is greater than any negative
number. As we distinguished between identities and equations
of condition in § 53, so in this discussion we observe that some
statements of inequality are true for any real value of the letters,
while others hold for particular values only. The former class
may be called unconditional inequalities, the latter conditional.

Thus a2 > — 1 is true for any real value of a and is unconditional, while
a; — 1 > 2 only when a; is greater than 3 and is consequently conditional.

The two inequalities <* > 5, g'> d are said to have the same
sense. Similarly, a <b, c <d have the same sense. The inequal-
ities a>h, c <id have a different sense.

Theorem I. Any positive number may he added to, subtracted
from, or midtiplied by both numbers of an inequality without
affecting the sense of the inequality.

Let a> b, that is, let a — b = k, where k is a positive number.
If m is a positive number, evidently

a ± m — (b ± m) = k,

or

a ±m> b±m.

Similarly,

ma — mb = mk,

or

ma > mb.

The other statements of the theorem are proved similarly.

Corollary. Terms may be transposed from one side of an
inequality to the other as in the case of equations.

269

Let a> b -{- G.

Subtract c from both sides of the inequality and we obtain by
Theorem I

a — c> h.

Theorem IL If the signs of both sides of an inequality are
changed, the sense of the inequality must be reversed, that is, the
> sign must be changed to <, or conversely.

Let a> b, that is, let a — b = k, where ^ is a positive number.

Then -a-\-b=-k,

or (— «)—(—&) = — 7c,

that is, by definition, — a < — b.

EXERCISES

Prove that the following identities are tnie for all real positive values of

the letters.

1. a2 + 62 > 2 ah.

Solution : (a — h)^ is always positive.

Thus a2 - 2 a6 + 62 = a2 + 62 _ 2 a6 is positive.

That is, a2 + 62 > 2 ah.

2. 3(a3 + 63)>a26 + a62.

3. a2 + 62 + c2 > a6 + ac + he.

4. (6 + c) (c + a) (a + 6) > 8 ahc.

5. (a + 6 + c) (pfi + 62 + c2) > 9a6c.

6. 62c2 + c2a2 + a262 > a6c (a + 6 + c).

7. 3(a8 4- 63 + c8) > (a + 6 + c) {ah + 6c + ca).

8. V(x + xi)2 + (y + 2/i)2 < Vx2 + 2/2 + y/x^^ + yi^.

9. If a2 + 62 = 1, x2 + y2 = 1^ prove that ax-\-hy< 1.

10. (a + 6 - c)2 + (a + c - 6)2 + (6 + c - a)2 > a6 + 6c 4- ca.

11. Show that the sum of any positive number (except 1) and its reciprocal
is greater than 2.

12. Prove that the arithmetical mean of two unequal positive numbers
always exceeds their geometrical mean.

INEQUALITIES 271

226. Conditional linear inequalities. If we wish to find the
values of x for which

ax -\- b < c, (1)

where a, h, and c are numbers and a is positive, we may find
such values by carrying out a process similar to that of solving a
linear equation in one variable.

By the corollary, § 225, we have from (1)

ax < c — b.

By Theorem I, § 225, x <-

227. Conditional quadratic inequalities. We have already
shown in § 116 that the quadratic expression ax^ -\- bx -\- c is posi-
tive or negative, when the equation

ax^ -{- bx -{- G = (1)

has imaginary or equal roots, according as a is positive or nega-
tive. If the equation has distinct real roots, the expression is
positive or negative for values between those roots according as
a is negative or positive. This we may express in tabular form
as follows, for all values of x excepting the roots of (1), for
which of course the expression vanishes.

a

62 _ 4 ac

dx^ -\- bx + c

+

-orO

Always +

-

-orO

Always —

+

+

- for X between roots, + for other values

-

+

+ for X between roots, - for other values

This enables us to answer immediately questions like the
following :

Example. For what values of cc is — 2 ic^+a; > — 3 ? By the corollary, § 225,
this is equivalent to the question, For what value ofxis —2x^-^x-\-S>0?

Here 62 — 4 ac = 1 -}- 24 = 25 is positive. The roots of the equation
— 2 a;2 4- a; + 3 = are ic = — 1, a; = f . Thus by our table this expression
is positive for all values of x between — 1 and |.

EXERCISES

For what values of x are the following inequalities valid ?

1. 2x-3>0.

3. _x-l>7.

- 9x 4 ^
3 7

7. .12x.+ .3<1.3.

9. 3<5x-2.

11. x2-8x + 22>6.
13. 2x2-3x>5.
15. 2x2 -4x< -2.
17. -3x2 + 2x<2.
19. 5x2-8x<l.
21. 3x2>3x-3.

2.

4x-7>l.

4.

- 3 X + 8 < 3.

6.

3 ^4^5

8.

3-4x>2.

10.

.s<i|-.

12.

x2 + 3x-2>l.

14.

-3x2-4x>8.

16.

3x2-9x>-6.

18.

- x2 + 6 X > 9.

20.

x2 < X - 1.

22.

3x>2x2-4.

CHAPTEE XXIII

VARIATION

228. General principles. The number x is said to vary directly
as the number y when the ratio of ic to 2/ is constant. , This we
symbolize by

i» oc 2/, or ^ = A^, (1)

where A; is a constant.

Thus if a man walks at a uniform speed, the distance that he
goes varies directly as the time. If the length of the altitude of
a triangle is given, the area of the triangle varies directly as the
base. The volume of a sphere varies directly as the cube of

The number x is said to vary inversely as the number y when
X varies directly as the reciprocal of y. Thus x varies inversely
as y when

05 oc -5 OY - — xy =^ Uy (2)

y i
y

where A: is a constant. Thus the speed of a horse might vary
inversely as the weight of his load. The length of time to do
a piece of work might vary inversely as the nmnber of laborers
employed.

The intensity of a light varies inversely as the square of the
distance of the light from the point of observation. If I repre-
sents the intensity of light and d the distance of the light from
the point of observation, we have

Zoc|. or| = ZcZ^ = A5, (3)

where A; is a constant. - ' -

273

274

The number x is said to vary jointly as y and z when it varies
directly as the product of y and z. Thus x varies jointly as y
and z when

X QC yz, or

yz

h

(4)

where Aj is a constant.

Thus a man's wages might vary jointly as the number of days
and the number of hours per day that he worked.

The number x is said to vary directly as y and inversely as z

when it varies directly with -• Thus the force of the attraction

of gravitation between two bodies varies directly as their masses
and inversely as the squares of their distances. If m represents
the masses of two bodies, d their distance, and G the force of
their attraction due to gravity, then

m
Goc^, or

= k.

(P)

EXERCISES

1. If a varies inversely as the square of 6, and if a = 2 when 6 = 3, what
is the value of a when 6 is 18 ?

Solution: By (3), a62 = k.

We can determine k by substituting a = 2, 6 = 3.

2 . 9 = A;.
18 = A:.

Then

a . (18)2 = 18,

or a = tV-

2. The volume of a sphere varies as the cube of its radius. A sphere of
radius 1 has a volume 4.19. What is the volume of a sphere of radius 3 ?

Solution: Let F represent the volume and r the radius of the sphere.
Then by (1),

\ = k.
4.19

Determine k by substituting.

Then

= k.

A; = 4. 19.

V V

-- = — = 4.19.
(3)8 27

F= 113.13.

VARIATION 275

3. ltxyxx + y, and a; = 1 when y = 1, find x when y = 8.

4. The area of a circle varies as the square of the radius. If a circle of
radius 1 has an area 3.14, find the area of a circle whose radius is 21.

5. Find the volume of a sphere whose radius is .2.
Hint. See exercise 2.

6. The volume of a circular cylinder varies jointly with the altitude and
the square of the radius of the base. A cylinder whose altitude and radius
are each 1 has a volume of 3. 14. Find the volume of a cylinder whose
altitude is 16 and whose radius is 3.

7. The weight of a body of a given material varies directly with its
volume. If a sphere of radius 1 inch weighs f of a pound, how much would
a ball of the same material weigh whose radius is 16 inches ?

8. The distance fallen by an object starting from rest varies as the square
of the time of falling. If a body falls 16 feet in 1 second, how far will it
fall in 6 seconds ?

9. A body falls from the top to the bottom of a cliff in 3| seconds. How
high is the cliff ?

10. A triangle varies in area jointly as its base and altitude. The area
of a triangle whose base and altitude are each 1 is ^. What is the area of a
triangle whose base is 16 and altitude 7?

11. If 6 men do a piece of work in 10 days, how long will it take 5 men
to do it?

12. If 3 men working 8 hours a day can finish a piece of work in 10 days,
how many days will 8 men require if they work 9 hours a day ?

13. An object is 30 feet from a light. To what point must it be moved in
order to receive (a) half as much light, (b) three times as much light ?

14. The weights of objects near the earth vary inversely as the squares
of their distances from the center of the earth. The radius of the earth is
about 4000 miles. If an object weighs 150 pounds on the surface of the
earth, how much would it weigh 6000 miles distant from the center ?

CHAPTEE XXIV
PROBABILITY

229. Illustration. If a bag contains 3 wlaite balls and 4 blact
balls, and 1 ball is taken out at random, what is the chance that
the ball drawn will be white ?

This question we may answer as follows : There are 7 balls in
the bag and we are as likely to get one as another. Thus a ball
may be drawn in 7 different ways. Of these 7 possible ways 3
will produce a white ball. Thus the chance that the ball drawn
will be white is 3 to 7, or f. The chance that a black ball will
be drawn is f .

230. General statement. It is plain that we may generalize
this illustration as follows : If an event may happen in ^ ways
and fail in q ways, each way being equally probable, the chance
or probability that it will happen in one of the ^ ways is

V

The chance that it will fail is

p + q

(1)

(2)

The sum of the chances of the event's happening and failing
is 1, as we observe by adding (1) and (2).

The odds in favor of the event are the ratio of the chance of
happening to the chance of failure. In this case the odds in
favor are

I. (3)

q

The odds against the event are -.

P
276

PKOBABILITi 277

EXERCISES

1. K the chance of an event's happening is ^^^ what are the odds in its
favor ?

v 1
Solution : By (1), -^— = — .

Hence lOjp = p + g,

or 9p = g,

«i 1
or - = - , which by (3) are the odds in favor.

q 9

2. From a pack of 52 cards 3 are missing. What is the chance that they
are all of a particular suit P

Solution : The number of combinations of 62 cards taken 3 at a time is

C52, 3 = — ' This represents p ■{■ q. The number of combinations of the

1 * 2 • 3 13 • 12 • 11

13 cards of any one suit taken 3 at a time is C13, 3 = ^ „ • This repre-
sentep,

13 • 12 • 11

^ p 1-2.3 13.12-11 11 11

Thus — =- — = = = ■ = .

p-\-q 62 » 61 • 60 62.61-60 17-60 860

1-2-3

3. What is the chance of throwing one and only one 6 in a single throw
of two dice ?

Solution : There are 36 possible ways for the two dice to fall. This repre-
sents p + g. Since a throw of two sixes is excluded there are 6 throws in
which each die would be a 6, that is, 10 in all in which a 6 appears. This
represents p.

P 10 6
Thus — - — = — = —

p + q 36 18

4. A bag contains 8 white and 12 black balls. What is the chance that a
ball drawn shall be (a) white, (b) black ?

5. A bag contains 4 red, 8 black, and 12 white balls. What is the chance
that a ball drawn shall be (a) red, (b) white, (c) not black ?

6. In the previous problem, if 3 balls are drawn, what is the chance that
(a) all are black, (b) 2 red and 1 white ?

7. What is the chance of throwing neither a 3 nor a 4 in a single throw
of one die ?

8. What is the chance in drawing a card from a pack that it be (a) an
ace, (b) a diamond, (c) a face card ?

9. Three cards are missing from a pack. What is the chance that they
are (a) of one color, (b) face cards, (c) aces ?

10. A coin is tossed twice. What is the chance that heads will fall
once?

11. The chance that an event will happen is f. What are the odds in its
favor ?

12. The odds against the occurrence of an event are |. What is the chance
of its happening ?

13.' What is the chance of throwing 10 with a single throw of two dice ?

14. A squad of 10 men stand in line. What is the chance that A and B
are next each other ?

15. What is the chance that in a game of whist a player has 6 trumps ?

16. What is the chance that in a game of whist a player holds 4 aces ?

CHAPTER XXV
SCALES OF NOTATION

231. General statement. The ordinary numbers with which
we are acquainted are expressed by means of powers of 10. Thus
263 = 2 . 10^ + 6 . 10^ + 3.
This is the common scale of notation, and 10 is called the radix
of the scale.

In a similar manner a number might be expressed in any scale
with any radix other than 10. If we take 6 as the radix, we shall
have as a 'number in this scale, for instance,
543 = 5 • 62 + 4 • 6 4- 3.
In this scale we need only and five digits to express every
positive integer.

In general, if r is the radix of a scale of notation, any positive
integer N will be denoted in this scale as follows :

N=ao7^ + a^r--^ + a^r^-^ + . . . + o,^. (1)

Theorem. Any positive . integer may he expressed in a scale
of notation of radix r.

Suppose we have a positive integer N. Let ?•" be the highest
power of r that is contained in N. Then

N = aoT- + iV'i,
where N^ is less than r". Suppose that on dividing iVj by r"-^ we
obtain ^^ ^ ^^^.„_i ^ ^^^

where N^ is less than r"~\

Then N = a^r^ + air«-i + N^.

Proceeding in this manner we obtain finally

N = aor"" -h 0^1^""^ H h a„,

where the a's are positive integers less than r, or perhaps zeros.

279

One observes that the symbol 10 indicates the radix in any
system. In this general scale we need a and r — 1 digits to
express every possible number.

232. Fundamental operations. In the four fundamental operar
tions in the common scale we carry and borrow 10 in computing.
In computing in a scale of radix 6, for instance, we should carry
and borrow 6. If the radix were r, we should carry or borrow r.

Thus let r = 6. Then 4 + 5=1-6+3=13. Similarly, 5-3=2. 6 + 3 = 23.
This is precisely analogous to our computation in the common scale, where, for
instance, we would have 9 + 8 = 1 • 10 + 7 = 17, or 6 • 7 = 4 - 10 + 2 = 42.

EXERCISES

Perform the following operations.

1. 2361 + 4253 + 2140 ; r = 7.

2361
4253
2140 -

12114

In this process, since 3 + 1=4 and is less than the radix, there is nothing to carry.
The next column gives 6 + 5 + 4= 15 =2-7+1, hence we write down 1 and carry 2. The
next column gives 3 + 2+1 + 2=8=1-7 + 1, hence we write down 1 and carry 1. Finally
we get 2 + 4 + 2 + 1 = 9=1-7 + 2, hence we write 12.

2. 4602-3714; r = 8.

4602
3714

Since we cannot take 4 from 2 we horrow one from the next place. Since the radix
is 8 this amounts to 8 units in the first place. We then subtract 4 from 8 + 2, which
leaves 6. In borrowing 1 from that digit is really reduced to 7 and the preceding digit
to 5 ; then subtracting 1 from 7 we get 6. Since we cannot take 7 from 5 we borrow 8
again and take 7 from 5 + 8= 13, which leaves 6. Since 1 has been borrowed from the 4
we see the subtraction is complete since 3-3=0.

3. 4321 . 432 ; r = 6.

4821

432

14142

24013

33334

4143222

In multiplying by 2 we have nothing to carry until we multiply 3 by 2. This gives
6=1-5 + 1. Hence we put down 1 and carry 1 to the product of 2 and 4. The addition of
the partial products is carried out as in exercise 1.

SCALES OF NOTATION 281

4. 32130 H- 43 ; r = 6.

43 1 32130 1 430
300
213
213
00

In making an estimate for the first figure in the quotient we divide 32 by 4, keeping
in mind that for this purpose 32=3-64-2. Thus 4 is contained in 20 just 5 times, but
since our entire divisor is 43 we take 4 as the first figure in the quotient. The multipli-
cations are of course performed as in exercise 3, excepting that here 6 is the radix.

5. 4361 + 2635 + 5542 ; r = 1. 6. 5344 - 3456 ; r = 7.
7. 2340 . 4101 ; r=6. 8. 6435 • 35 ; r = 7.

9. 2003455 - 403 ; r = 6. 10. 344032 - 321 ; r = 5.

11. 534401 - 443524 ; r = 6. 12. 425 + 254 + 542 + 452 ; r=6.

233. Change of scale. If we have a number in the scale of
radix r, we may find the expression for that number in the com-
mon scale by writing the number in form (1), § 231, and carrying
out the indicated operations..

Example. Convert 4635, where r = 7, into the ordinary scale.
4635 = 4- 73 + 6. 72 + 3-7+5
= 4 • 343 + 6 . 40 + 3 . 7 + 5
= 1692.

If we have a number in the common scale, we may express it
in the scale with radix r as follows : If the number is N, we have
to determine the integers aQ, a^, • • , a„ in the expression

N = ao7^ + «^i^""^ H h a^-iT + ««• (1)

Divide (1) by r. We have

:^= aor"-i + a,r-' H- a^,_, + - = iV' +-;
r r r

that is, the remainder a^ of this division is the last digit in the
expression desired.

Divide N' by r and we obtain

— = JSJ-" = a^r^-^ + a^r^-^ H -|- ^^^^ ;

r r

that is, the remainder from this division is the next to the last
digit in the desired expression. Proceeding in this way we obtain
all the digits fl^„, c^„ _ i, • • • , ^i, «o-

Example. Express 37496 in the scale with radix 7.

7) 37496

7 ) 5356 remainder 4

1 ) 765 remainder 1

• 7 ) 109 remainder 2

7 ) 15 remainder 4

7 )2 remainder 1

remainder 2

The number in scale r = 7 is 214214.

To change a number from any scale rj to any other scale r^j
we may first change the number to the scale of 10 and then by
the process just given to the scale r^. The process indicated in
the preceding example may be employed directly to change from
any scale to any other, provided the division is carried out in the
scale in which the number is given. One of these methods may
be used to check the other.

Example. Change 34503 from scale r = 6 to one in which r = 9.

34503 = 3. 6* + 4- 63 +5. 62 + 3 = 4935 in scale of 10. ♦

9 )4935

9 ) 548 remainder 3

9 )60 remainder 8

9)6 remainder 6

remainder 6

Thus 34503 in scale of 6 becomes 6683 in scale of 9.

Check: 9)34503

9)2312 remainder 3 ^" carrying out this division it must

■ X • A Q ^® kept in mind that the dividends are in

92140 remamder 8 g^^jg ^f g^ ^l^jle ^^Q remainders are to be

9) 10 remainder 6 in scale of 9.
remainder 6

234. Fractions. In the ordinary notation we express fractional
numbers by digits following the decimal point. This notation
may also be used in a scale with any radix.

SCALES OF NOTATION 283

Thus the expression .5421 stands for

10 102 ^ 10« 10*
in the common scale.

In the scale with radix r it stands for

ry% ry%** n%9 niT

The process of changing the scale for fractions is performed
in accordance with the same principles as are employed in the
change of scale for integers. The following examples suffice to
illustrate it.

Example 1. Express .5421 in the scale of 6 as a decimal fraction.

.6421 = 5 + 1+1+1
6 62 68 64

5.63 + 4.62+2-6+1 1237 ^,,,
6* 1296

Example 2. Express .439 as a fraction for radix 6.

Let .439 = ^ + A + l + l + ....

6 62 68 6*

Multiplying by 6, 2.634 = a + ^ + ^ + ^ + • . . .

b o-' 6'

Thus a = 2 and we have .634 = - + — f- 1 .

6 ^ 62 68

c d
Multiplying by 6, 3.804 = 6 + - + — + ... .

6 62

c d
Thus 6 = 3 and we have .804 = - + — + ••. .

6 62^

Multiplying by 6, 4.824 = c + - + •• ..

6

d
Thus c = 4 and we have .824 = - + ....

6

Multiplying by (T, 4.944 = d + • . ..

Thus d = 4.

The fraction in scale of radix 6 is then .2344 • • «„

EXERCISES

1. Express the following as decimal fractions,
(a) .374; r = 8. (b) .4352; r = 6.
(c) .2231 ; r = 4. (d) .2001 ; r = 3.

2. Express the decimal fraction .296 as a radix fraction for r = 5.

3. Express the decimal fraction .3405 as a radix fraction for r = Q.

34

4. Express — as a radix fraction for r = 4.

^ 128

5. Express as a radix fraction for r = 5.

^ 626

6. In what scale is 42 expressed as 1120?

Solution : We seek r where

r3 + r2 + 2 r = 42.

This is equivalent to finding a positive integral root of the equation

r3 + r2 + 2 r - 42 = 0.
By synthetic division,

1 + 1 + 2 - 42(2
+ 2 + 6 + 16
+ 3 4- 8-26
1 + 1+ 2-4213
+ 3 + 12 + 42
+ 4+14
Thus 3 is the value sought.

Check : 3^ + 32 + 2 • 3 = 27 + 9 + 6 = 42.

7. In what scale is 2704 denoted by 20304 ?

8. In what scale is 256 denoted by 10000 ?

9. In what scale is .1664 denoted by .0404 ?
10. Show that 1331 is a perfect cube.

235. Duodecimals. We may apply some of the foregoing
processes to mensuration. If we take one foot as a unit and the
radix as 12, we may express distances in the so-called duodecimal
notation. Thus 2 ft. 6 in. is represented in the duodecimal scale
by 2.6. Since in a scale of radix r we need ?• — 1 symbols, we
will let 10 = ^ and 11 = e. Thus 21 ft. 10 in. would be expressed
in duodecimal notation as 19.^. We may now find areas and vol-
umes in this notation much more readily than by the usual method
of converting all distances to inches.

SCALES OF NOTATION 285

Example. Multiply 8 ft. 3 in. by 3 ft. 10 in. We multiply 8 • 3 by 3.i

8.3 in the duodecimal scale. To convert the result to square

3-t feet and square inches we must keep in mind that 27.76

6t6 ^ 2 . 12 + 7 + tV + xf^ = 31 sq. ft. 90 sq. in., since 144

— — square inches equal one square foot.

This example suggests the following method of multiplying
distances :

EuLE. Express the distances in duodecimal notation with the
foot as a unit.

Multiply in the scale for which r = 12.

In the product change the part on the left of the point from
duodecimal to decimal scale.

Multiply the digit following the point hy 12^ and add to the
last figure to obtain the square inches in the result.

EXERCISES

1. Multiply the following :

(a) 13 ft. 4 in. by 67 ft. 11 in.

Solution: 11.4

57.e
1028
794

568

636.68 = 905 sq. ft. 80 sq. in.

(b) 10 ft. 6 in. by 12 ft. 2 in.

(c) 8 ft. 4 in. by 11 ft. 11 in.

(d) 23 ft. 6 in. by 47 ft. 8 in.

(e) 41 ft. 6 in. by 36 ft. 1 in.

2. What is the area of a room 16 ft. 2 in. by 10 ft. 3 in. ?

3. What is the area of a walk 60 ft. 6 in. by 3 ft. 3 in. ?

4. What is the area of a city lot 62 ft. 6 in. by 163 ft. 7 in. ?

^4

UNIVERSITY OF CALIFORNIA LIBRARY
BERKELEY

This book is DUE on the last date stamped below.

SEP 7 'S5h Ll.
100ct'55CT

,5^ov'56?W
RECD LD
NOV 1 ^9^^

«?Jul'57lVlH

JUL 3 wa/

RECD LCi

DEC 18 135f

LD 21-100ot-7,'52 (A2528s16)476

AUG as* ^^;^«

UNIVERSITY OF CALIFORNIA UBRARY

;':f:^

```